Anand Classes provides high-quality NCERT Solutions for the Miscellaneous Exercise of Chapter 7 Integrals for Class 12 Mathematics, helping students build strong conceptual understanding through step-by-step explanations and well-structured solutions. These Set-3 notes follow the latest NCERT syllabus and are designed to support effective revision, doubt clarification, and board exam preparation. Click the print button to download study material and notes.
NCERT Question.21 : Evaluate the integral
$$\displaystyle \int \frac{x^{2}+x+1}{(x+1)^{2}(x+2)}\;dx$$
Solution
$$\displaystyle \int \frac{x^{2}+x+1}{(x+1)^{2}(x+2)}\;dx$$
Use partial fractions. Assume
$$
\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C}{x+2}.
$$
Multiplying through by $(x+1)^{2}(x+2)$ gives
$$
x^{2}+x+1=A(x+1)(x+2)+B(x+2)+C(x+1)^{2}.
$$
Expanding and comparing coefficients yields the system
$$
\begin{cases}
A+C=1,\\
3A+B+2C=1,\\
2A+2B+C=1.
\end{cases}
$$
Solving gives
$$
A=-2,\quad B=1,\quad C=3.
$$
Thus
$$
\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}=-\frac{2}{x+1}+\frac{1}{(x+1)^{2}}+\frac{3}{x+2}.
$$
Integrate termwise:
$$
\int \frac{x^{2}+x+1}{(x+1)^{2}(x+2)}\;dx
=\int\Bigl(-\frac{2}{x+1}+\frac{1}{(x+1)^{2}}+\frac{3}{x+2}\Bigr)\;dx$$
$$\int \frac{x^{2}+x+1}{(x+1)^{2}(x+2)}\;dx=-2\ln|x+1|-\frac{1}{x+1}+3\ln|x+2|+C$$
Final Result
$$
\boxed{\;\displaystyle -2\ln|x+1|+3\ln|x+2|-\frac{1}{x+1}+C\;}
$$
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NCERT Question.22 : Evaluate the integral
$$\displaystyle \int \tan^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)\;dx$$
Solution
$$
\displaystyle \int \tan^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)\;dx
$$
Put
$$
x=\cos 2\theta\qquad\Rightarrow\qquad dx=-2\sin 2\theta\;d\theta=-4\sin\theta\cos\theta\;d\theta.
$$
Note
$$
\sqrt{\frac{1-x}{1+x}}=\sqrt{\frac{1-\cos 2\theta}{1+\cos 2\theta}}=\sqrt{\frac{2\sin^{2}\theta}{2\cos^{2}\theta}}=\tan\theta
$$
so $\tan^{-1}\bigl(\sqrt{\dfrac{1-x}{1+x}}\bigr)=\theta$. Hence
$$
\int \tan^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)\;dx
=\int \theta\;(-4\sin\theta\cos\theta)\;d\theta
=-4\int \theta\sin\theta\cos\theta\;d\theta.
$$
Use $\sin\theta\cos\theta=\tfrac{1}{2}\sin 2\theta$ to get
$$
-4\int \theta\sin\theta\cos\theta\;d\theta
=-2\int \theta\sin 2\theta\;d\theta.
$$
Integrate by parts with $u=\theta,\;dv=\sin 2\theta\;d\theta$ so $du=d\theta,\;v=-\tfrac{1}{2}\cos 2\theta$:
$$
-2\int \theta\sin 2\theta\;d\theta
=-2\Bigl(\theta\cdot\bigl(-\tfrac{1}{2}\cos 2\theta\bigr)-\int \bigl(-\tfrac{1}{2}\cos 2\theta\bigr)\;d\theta\Bigr).
$$
So
$$
-2\int \theta\sin 2\theta\;d\theta
=\theta\cos 2\theta-\tfrac{1}{2}\sin 2\theta +C.
$$
Return to $x$. Since $\cos 2\theta=x$ and $\sin 2\theta=\sqrt{1-x^{2}}$ (taking the principal sign on $-1<x<1$), and $\theta=\tan^{-1}\bigl(\sqrt{\tfrac{1-x}{1+x}}\bigr)$ or equivalently $\theta=\tfrac{1}{2}\cos^{-1}x$, we may write the antiderivative in simpler form:
$$\theta\cos 2\theta-\tfrac{1}{2}\sin 2\theta +C=\displaystyle \frac{x}{2}\cos^{-1}x-\frac{1}{2}\sqrt{1-x^{2}}+C$$
$$\displaystyle \int \tan^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)\;dx=\displaystyle \frac{x}{2}\cos^{-1}x-\frac{1}{2}\sqrt{1-x^{2}}+C$$
Final Result
$$
\boxed{\displaystyle \;\frac{x}{2}\cos^{-1}x-\frac{1}{2}\sqrt{1-x^{2}}+C\;}$$
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NCERT Question.23 : Evaluate the integral
$$\int \frac{\sqrt{x^{2}+1}}{x^{4}}\Bigl[\log(x^{2}+1)-2\log x\Bigr]\;dx$$
Solution
$$\int \frac{\sqrt{x^{2}+1}}{x^{4}}\Bigl[\log(x^{2}+1)-2\log x\Bigr]\;dx$$
First simplify the logarithm:
$$
\log(x^{2}+1)-2\log x=\log\Bigl(1+\frac{1}{x^{2}}\Bigr).
$$
Hence the integral is
$$
I=\int \frac{\sqrt{x^{2}+1}}{x^{4}}\;\log\Bigl(1+\frac{1}{x^{2}}\Bigr)\;dx.
$$
Put $u=\dfrac{1}{x}$ (so $x=\dfrac{1}{u},\;dx=-\dfrac{1}{u^{2}}\;du$). Then
$$
\sqrt{x^{2}+1}=\frac{\sqrt{1+u^{2}}}{u},\qquad
\frac{\sqrt{x^{2}+1}}{x^{4}}=u^{3}\sqrt{1+u^{2}}.
$$
Thus
$$
I=-\int u\sqrt{1+u^{2}}\;\log(1+u^{2})\;du.
$$
Now put $t=1+u^{2}$ so $dt=2u\;du$ and $u\;du=\frac{1}{2}\;dt$. The integral becomes
$$I=-\frac{1}{2}\int t^{1/2}\log t\;dt$$
Integrate by parts with $A=\log t,\ dB=t^{1/2}dt$ (so $dA=\dfrac{dt}{t},\ B=\dfrac{2}{3}t^{3/2}$):
$$
I=-\frac{1}{2}\Bigl[\dfrac{2}{3}t^{3/2}\log t-\dfrac{2}{3}\int t^{1/2}dt\Bigr]$$
$$I=-\frac{1}{2}\cdot\dfrac{2}{3}t^{3/2}\Bigl(\log t-\frac{2}{3}\Bigr).
$$
Hence
$$
I=-\frac{1}{3}t^{3/2}\Bigl(\log t-\frac{2}{3}\Bigr)+C.
$$
Return to $u$ and then to $x$. Since $t=1+u^{2}=1+\dfrac{1}{x^{2}}$ and
$$
t^{3/2}=\Bigl(1+\frac{1}{x^{2}}\Bigr)^{3/2}=\frac{(x^{2}+1)^{3/2}}{x^{3}}
$$
we obtain the final form
$$
\boxed{\displaystyle I
=-\frac{1}{3}\frac{(x^{2}+1)^{3/2}}{x^{3}}\left[\log\Bigl(1+\frac{1}{x^{2}}\Bigr)-\frac{2}{3}\right]+C.}
$$
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NCERT Question.24 : Evaluate the integral
$$
\displaystyle I=\int_{\pi/2}^{\pi} e^{x}\left(\frac{1-\sin x}{1-\cos x}\right)\;dx
$$
Solution
$$
\displaystyle I=\int_{\pi/2}^{\pi} e^{x}\left(\frac{1-\sin x}{1-\cos x}\right)\;dx
$$
Use the half-angle identities :
$$
1-\cos x=2\sin^{2}\frac{x}{2}\; , \qquad
\sin x=2\sin\frac{x}{2} 2\cos\frac{x}{2}.
$$
Hence
$$
\frac{1-\sin x}{1-\cos x}=\frac{1}{2}\csc^{2}\frac{x}{2}-\cot\frac{x}{2}.
$$
Set
$$
f(x)=-\cot\frac{x}{2}\;, \qquad f'(x)=\frac{1}{2}\csc^{2}\frac{x}{2}\;
$$
so the integrand is $f'(x)+f(x)$. Therefore
$$
e^{x}\left(\frac{1-\sin x}{1-\cos x}\right)=e^{x}\bigl(f(x)+f'(x)\bigr)
$$
and an antiderivative is $e^{x}f(x)$ because
$$
\frac{d}{dx}\bigl(e^{x}f(x)\bigr)=e^{x}f(x)+e^{x}f'(x).
$$
$$
\displaystyle I=\int_{\pi/2}^{\pi} e^{x}\left(\frac{1-\sin x}{1-\cos x}\right)\;dx=\int_{\pi/2}^{\pi} e^{x}\left(\frac{1}{2}\csc^{2}\frac{x}{2}-\cot\frac{x}{2}\right)\;dx
$$
Thus
$$
I=\Bigl[e^{x}\bigl(-\cot\frac{x}{2}\bigr)\Bigr]_{x=\pi/2}^{x=\pi}.
$$
Evaluate the boundary terms:
$$
\cot\frac{\pi}{2}=0\;, \qquad \cot\frac{\pi}{4}=1\;
$$
so
$$
I=\bigl(e^{\pi}\cdot 0\bigr)-\bigl(e^{\pi/2}\cdot(-1)\bigr)=e^{\pi/2}.
$$
Final Result
$$
\boxed{\displaystyle I=e^{\pi/2}}
$$
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NCERT Question.25 : Evaluate the integral
$$\displaystyle \int_{0}^{\pi/4}\frac{\sin x\;\cos x}{\cos^{4}x+\sin^{4}x}\;dx$$
Solution
$$\displaystyle \int_{0}^{\pi/4}\frac{\sin x\;\cos x}{\cos^{4}x+\sin^{4}x}\;dx$$
Divide numerator and denominator by $\cos^{4}x$ to rewrite the integrand in terms of $\tan x$ and $\sec x$:
$$
\frac{\sin x\;\cos x}{\cos^{4}x+\sin^{4}x}
=\frac{\tan x\;\cos^{2}x}{\cos^{4}x(1+\tan^{4}x)}
=\frac{\tan x\;\sec^{2}x}{1+\tan^{4}x}.
$$
Put
$$
u=\tan^{2}x\qquad\Rightarrow\qquad du=2\tan x\;\sec^{2}x\;dx,
$$
so
$$
\tan x\;\sec^{2}x\;dx=\frac{1}{2}\;du,
\qquad
1+\tan^{4}x=1+u^{2}.
$$
Thus the integral becomes
$$
\int_{0}^{\pi/4}\frac{\tan x\;\sec^{2}x}{1+\tan^{4}x}\;dx
=\frac{1}{2}\int_{u(0)}^{u(\pi/4)}\frac{du}{1+u^{2}}
=\frac{1}{2}\int_{0}^{1}\frac{du}{1+u^{2}}.
$$
Evaluate:
$$
\frac{1}{2}\bigl[\tan^{-1}u\bigr]_{0}^{1}
=\frac{1}{2}\left(\frac{\pi}{4}-0\right)
=\frac{\pi}{8}.
$$
Final Result
$$
\boxed{\displaystyle \frac{\pi}{8}}
$$
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NCERT Question.26 : Evaluate the integral
$$\displaystyle I=\int_{0}^{\dfrac{\pi}{2}}\frac{\cos^{2}x}{\cos^{2}x+4\sin^{2}x}\;dx$$
Solution
Let
$$I=\int_{0}^{\dfrac{\pi}{2}}\frac{\cos^{2}x}{\cos^{2}x+4\sin^{2}x}\;dx.$$
Use $\sin^{2}x=1-\cos^{2}x$ to rewrite the denominator:
$$\cos^{2}x+4\sin^{2}x=\cos^{2}x+4(1-\cos^{2}x)=4-3\cos^{2}x.$$
Thus
$$I=\int_{0}^{\dfrac{\pi}{2}}\frac{\cos^{2}x}{4-3\cos^{2}x}\;dx.$$
Write the integrand as
$$\frac{\cos^{2}x}{4-3\cos^{2}x}
=\frac{1}{-3}\cdot\frac{4-3\cos^{2}x-4}{4-3\cos^{2}x}$$
$$\frac{\cos^{2}x}{4-3\cos^{2}x}=-\frac{1}{3}\cdot\frac{4-3\cos^{2}x}{4-3\cos^{2}x}+\frac{1}{3}\cdot\frac{4}{4-3\cos^{2}x}.$$
So
$$I=-\frac{1}{3}\int_{0}^{\dfrac{\pi}{2}}1\;dx+\frac{1}{3}\int_{0}^{\dfrac{\pi}{2}}\frac{4}{4-3\cos^{2}x}\;dx.$$
The first integral is straightforward:
$$-\frac{1}{3}\int_{0}^{\dfrac{\pi}{2}}1\;dx=-\frac{1}{3}\left[x\right]_{0}^{\frac{\pi}{2}}=-\frac{\pi}{6}.$$
For the second integral, express in terms of $\tan x$ and $\sec^{2}x$:
$$\frac{4}{4-3\cos^{2}x}=\frac{4\sec^{2}x}{4\sec^{2}x-3}.$$
Thus
$$\frac{1}{3}\int_{0}^{\dfrac{\pi}{2}}\frac{4}{4-3\cos^{2}x}\;dx=\frac{1}{3}\int_{0}^{\dfrac{\pi}{2}}\frac{4\sec^{2}x}{4\sec^{2}x-3}\;dx.$$
Factor $4\sec^{2}x$ in numerator and use $\sec^{2}x=1+\tan^{2}x$:
$$\frac{4\sec^{2}x}{4\sec^{2}x-3}=\frac{4\sec^{2}x}{4(1+\tan^{2}x)-3}=\frac{4\sec^{2}x}{1+4\tan^{2}x}.$$
So
$$\frac{1}{3}\int_{0}^{\dfrac{\pi}{2}}\frac{4\sec^{2}x}{4\sec^{2}x-3}\;dx=\frac{4}{3}\int_{0}^{\dfrac{\pi}{2}}\frac{\sec^{2}x}{1+4\tan^{2}x}\;dx.$$
Make the substitution $t=2\tan x\;\Rightarrow\;dt=2\sec^{2}x\;dx$, so $\sec^{2}x\;dx=\dfrac{dt}{2}$. When $x=0$, $t=0$, when $x=\dfrac{\pi}{2}$, $t\to\infty$. Therefore
$$\frac{4}{3}\int_{0}^{\dfrac{\pi}{2}}\frac{\sec^{2}x}{1+4\tan^{2}x}\;dx
=\frac{4}{3}\int_{0}^{\infty}\frac{1}{2}\cdot\frac{dt}{1+t^{2}}
=\frac{2}{3}\int_{0}^{\infty}\frac{dt}{1+t^{2}}.$$
Evaluate the arctangent integral:
$$\frac{2}{3}\int_{0}^{\infty}\frac{dt}{1+t^{2}}=\frac{2}{3}\left[\tan^{-1}t\right]_{0}^{\infty}=\frac{2}{3}\cdot\frac{\pi}{2}=\frac{\pi}{3}.$$
Combine both parts:
$$I=-\frac{\pi}{6}+\frac{\pi}{3}=\frac{\pi}{6}.$$
Final Result
$$\boxed{\;I=\dfrac{\pi}{6}\;}$$
NCERT Question.27 : Evaluate the integral
$$\displaystyle I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\sin x+\cos x}{\sqrt{\sin2x}}\;dx$$
Solution
$$\displaystyle I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\sin x+\cos x}{\sqrt{\sin2x}}\;dx$$
Use the substitution
$$t=\sin x-\cos x\quad\Rightarrow\quad dt=(\cos x+\sin x)\;dx.$$
Also note
$$(\sin x-\cos x)^{2}=1-\sin2x\quad\Rightarrow\quad \sin2x=1-t^{2}.$$
Hence the integrand becomes
$$\frac{\sin x+\cos x}{\sqrt{\sin2x}}\;dx=\frac{dt}{\sqrt{1-t^{2}}}.$$
Compute the limits. When $x=\dfrac{\pi}{6}$,
$$t=\sin\frac{\pi}{6}-\cos\frac{\pi}{6}=\frac{1}{2}-\frac{\sqrt{3}}{2}=\frac{1-\sqrt{3}}{2}=-\frac{\sqrt{3}-1}{2}.$$
When $x=\dfrac{\pi}{3}$,
$$t=\sin\frac{\pi}{3}-\cos\frac{\pi}{3}=\frac{\sqrt{3}}{2}-\frac{1}{2}=\frac{\sqrt{3}-1}{2}.$$
Therefore
$$I=\int_{-\frac{\sqrt{3}-1}{2}}^{\frac{\sqrt{3}-1}{2}}\frac{dt}{\sqrt{1-t^{2}}}
=\left[\sin^{-1}t\right]_{-\frac{\sqrt{3}-1}{2}}^{\frac{\sqrt{3}-1}{2}}$$
$$I =\sin^{-1}\left(\frac{\sqrt{3}-1}{2}\right)-\sin^{-1}\left(-\frac{\sqrt{3}-1}{2}\right).$$
Since $\sin^{-1}(-a)=-\sin^{-1}(a)$, this simplifies to
$$I=2\sin^{-1}\left(\frac{\sqrt{3}-1}{2}\right).$$
Final Result
$$\boxed{\;I=2\sin^{-1}\left(\frac{\sqrt{3}-1}{2}\right)\;}$$
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NCERT Question.28 : Evaluate the integral
$$I=\int_{0}^{1}\frac{dx}{\sqrt{1+x}-\sqrt{x}}$$
Solution
$$I=\int_{0}^{1}\frac{dx}{\sqrt{1+x}-\sqrt{x}}$$
Multiply by the conjugate of the denominator:
$$
\frac{1}{\sqrt{1+x}-\sqrt{x}}
=\frac{\sqrt{1+x}+\sqrt{x}}{(\sqrt{1+x}-\sqrt{x})(\sqrt{1+x}+\sqrt{x})}
=\sqrt{1+x}+\sqrt{x}.
$$
Hence
$$
I=\int_{0}^{1}\big(\sqrt{1+x}+\sqrt{x}\big)\;dx
=\int_{0}^{1}(1+x)^{\tfrac{1}{2}}\;dx+\int_{0}^{1}x^{\tfrac{1}{2}}\;dx.
$$
Evaluate each integral:
$$
\int(1+x)^{\tfrac{1}{2}}\;dx=\frac{2}{3}(1+x)^{\tfrac{3}{2}},\qquad
\int x^{\tfrac{1}{2}}\;dx=\frac{2}{3}x^{\tfrac{3}{2}}.
$$
So
$$
I=\frac{2}{3}\big[(1+x)^{\tfrac{3}{2}}\big]_{0}^{1}+\frac{2}{3}\big[x^{\tfrac{3}{2}}\big]_{0}^{1}
=\frac{2}{3}\big(2^{\tfrac{3}{2}}-1\big)+\frac{2}{3}\big(1-0\big)
=\frac{2}{3}\cdot 2^{\tfrac{3}{2}}.
$$
Since $2^{\tfrac{3}{2}}=2\sqrt{2}$, we get
Final Result
$$\boxed{I=\dfrac{4\sqrt{2}}{3}}$$
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NCERT Question.29 : Evaluate the integral
$$\displaystyle I=\int_{0}^{\dfrac{\pi}{4}}\frac{\sin x+\cos x}{9+16\sin2x}\;dx$$
Solution
$$\displaystyle I=\int_{0}^{\dfrac{\pi}{4}}\frac{\sin x+\cos x}{9+16\sin2x}\;dx$$
Let
$$t=\sin x-\cos x\quad\Rightarrow\quad dt=(\cos x+\sin x)\;dx.$$
Use the identity
$$(\sin x-\cos x)^{2}=\sin^{2}x+\cos^{2}x-2\sin x\cos x=1-\sin2x$$
$$\sin2x=1-t^{2}.$$
Under this substitution the limits become
$$x=0\Rightarrow t=\sin0-\cos0=0-1=-1, $$
$$x=\frac{\pi}{4}\Rightarrow t=\sin\frac{\pi}{4}-\cos\frac{\pi}{4}=0$$
Thus the integral transforms to
$$I=\int_{-1}^{0}\frac{dt}{9+16(1-t^{2})}
=\int_{-1}^{0}\frac{dt}{25-16t^{2}}.$$
Write $25-16t^{2}=5^{2}-(4t)^{2}$.
The antiderivative for $\dfrac{1}{a^{2}-b^{2}t^{2}}$ is $\dfrac{1}{2ab}\ln\dfrac{a+bt}{a-bt}$.
Therefore
$$\int\frac{dt}{25-16t^{2}}
=\frac{1}{2\cdot5\cdot4}\ln\frac{5+4t}{5-4t}
=\frac{1}{40}\ln\frac{5+4t}{5-4t}.$$
Evaluate from $t=-1$ to $t=0$:
$$
I=\frac{1}{40}\left[\ln\frac{5+4t}{5-4t}\right]_{-1}^{0}
=\frac{1}{40}\left(\ln\frac{5+0}{5-0}-\ln\frac{5-4}{5+4}\right)$$
$$I=\frac{1}{40}\left(0-\ln\frac{1}{9}\right)=\frac{1}{40}\ln 9$$
Final Result
$$\boxed{\;I=\dfrac{1}{40}\ln 9\;}$$
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NCERT Question.30 : Evaluate the integral
$$\displaystyle I=\int_{0}^{\dfrac{\pi}{2}}\sin2x \;\tan^{-1}(\sin x)\;dx$$
Solution
$$\displaystyle I=\int_{0}^{\dfrac{\pi}{2}}\sin2x \;\tan^{-1}(\sin x)\;dx$$
Let $t=\sin x\Rightarrow dt=\cos x\;dx$. Note $\sin2x=2\sin x\cos x$. Therefore
$$
I=\int_{0}^{\dfrac{\pi}{2}}2\sin x\cos x\;\tan^{-1}(\sin x)\;dx
=2\int_{0}^{1}t\;\tan^{-1}t\;dt.
$$
Compute $ \displaystyle \int t\;\tan^{-1}t\;dt $ by integration by parts. Take
$u=\tan^{-1}t\Rightarrow du=\dfrac{dt}{1+t^{2}}$ and $dv=t\;dt\Rightarrow v=\dfrac{t^{2}}{2}$. Then
$$
\int t\;\tan^{-1}t\;dt=\frac{t^{2}}{2}\tan^{-1}t-\int\frac{t^{2}}{2(1+t^{2})}\;dt.
$$
Write $\dfrac{t^{2}}{1+t^{2}}=1-\dfrac{1}{1+t^{2}}$ to get
$$
\int t\;\tan^{-1}t\;dt=\frac{t^{2}}{2}\tan^{-1}t-\frac{1}{2}\int\Big(1-\frac{1}{1+t^{2}}\Big)\;dt$$
$$\int t\;\tan^{-1}t\;dt=\frac{t^{2}}{2}\tan^{-1}t-\frac{1}{2}\Big(t-\tan^{-1}t\Big)$$
Simplify:
$$
\int t\;\tan^{-1}t\;dt=\frac{1}{2}(t^{2}+1)\tan^{-1}t-\frac{t}{2}.
$$
Evaluate from $0$ to $1$:
$$
\int_{0}^{1}t\;\tan^{-1}t\;dt=\left[\frac{1}{2}(t^{2}+1)\tan^{-1}t-\frac{t}{2}\right]_{0}^{1}
=\frac{1}{2}\cdot 2\cdot\frac{\pi}{4}-\frac{1}{2}
=\frac{\pi}{4}-\frac{1}{2}.
$$
Thus
$$
I=2\Big(\frac{\pi}{4}-\frac{1}{2}\Big)=\frac{\pi}{2}-1.
$$
Final Result
$$\boxed{\;I=\dfrac{\pi}{2}-1\; }$$
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