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NCERT Question 13: Evaluate the integral
$$\int \frac{2}{(1-x)(1+x^2)} \; dx$$
Solution
$$\int \frac{2}{(1-x)(1+x^2)} \; dx$$
Decompose using partial fractions:
$$
\frac{2}{(1-x)(1+x^2)} = \frac{A}{1-x} + \frac{Bx+C}{1+x^2}
$$
Multiply through by $(1-x)(1+x^2)$:
$$
2 = A(1+x^2) + (Bx+C)(1-x)
$$
Expand:
$$
2 = A + Ax^2 + Bx – Bx^2 + C – Cx
$$
Combine like terms and equate coefficients:
$$
x^2: A-B = 0 \quad \Rightarrow B = A
$$
$$
x: B-C = 0 \quad \Rightarrow C = B
$$
$$
\text{constant term: } A + C = 2
$$
Solving:
$$
A = 1, \quad B = 1, \quad C = 1
$$
So:
$$
\frac{2}{(1-x)(1+x^2)} = \frac{1}{1-x} + \frac{x+1}{1+x^2}
$$
Step 1: Integrate each term
$$
\int \frac{2}{(1-x)(1+x^2)} \; dx = \int \frac{1}{1-x} \; dx + \int \frac{x}{1+x^2} \; dx + \int \frac{1}{1+x^2} \; dx
$$
$$
\int \frac{2}{(1-x)(1+x^2)} \; dx = -\int \frac{1}{x-1} \; dx + \frac{1}{2} \int \frac{2x}{1+x^2} \; dx + \int \frac{1}{1+x^2} \; dx
$$
$$
\int \frac{2}{(1-x)(1+x^2)} \; dx = -\ln|x-1| + \frac{1}{2} \ln|1+x^2| + \tan^{-1}x + C
$$
Final Answer
$$
\boxed{-\ln|x-1| + \frac{1}{2} \ln|1+x^2| + \tan^{-1}x + C}
$$
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NCERT Question 14: Evaluate the integral
$$\int \frac{3x-1}{(x+2)^2} \; dx $$
Solution
$$\int \frac{3x-1}{(x+2)^2} \; dx $$
Decompose using partial fractions:
$$
\frac{3x-1}{(x+2)^2} = \frac{A}{x+2} + \frac{B}{(x+2)^2}
$$
Multiply through by $(x+2)^2$:
$$
3x-1 = A(x+2) + B
$$
Equate coefficients:
$$
x: A = 3
$$
$$
\text{constant term: } 2A + B = -1 \quad \Rightarrow B = -7
$$
So:
$$
\frac{3x-1}{(x+2)^2} = \frac{3}{x+2} – \frac{7}{(x+2)^2}
$$
Step 1: Integrate each term
$$
\int \frac{3x-1}{(x+2)^2} \; dx = \int \frac{3}{x+2} \; dx – \int \frac{7}{(x+2)^2} \; dx
$$
$$
\int \frac{3x-1}{(x+2)^2} \; dx = 3 \ln|x+2| – 7 \int (x+2)^{-2} \; dx
$$
$$
\int \frac{3x-1}{(x+2)^2} \; dx = 3 \ln|x+2| – 7 \left(-\frac{1}{x+2}\right) + C
$$
$$
\int \frac{3x-1}{(x+2)^2} \; dx = 3 \ln|x+2| + \frac{7}{x+2} + C
$$
Final Answer
$$
\boxed{3 \ln|x+2| + \frac{7}{x+2} + C}
$$
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NCERT Question 15: Evaluate the integral
$$ \int \frac{1}{x^4-1} \; dx $$
Solution
$$ \int \frac{1}{x^4-1} \; dx $$
Factorize the denominator:
$$
x^4-1 = (x^2-1)(x^2+1) = (x-1)(x+1)(x^2+1)
$$
Use partial fractions:
$$
\frac{1}{x^4-1} = \frac{A}{x+1} + \frac{B}{x-1} + \frac{Cx+D}{x^2+1}
$$
Multiply through by $(x-1)(x+1)(x^2+1)$:
$$
1 = A(x-1)(x^2+1) + B(x+1)(x^2+1) + (Cx+D)(x^2-1)
$$
Expand and equate coefficients:
$$
x^3: A+B+C = 0
$$
$$
x^2: -A+B+D = 0
$$
$$
x^1: A+B-C = 0
$$
$$
\text{constant: } -A+B-D = 1
$$
Solving, we get:
$$
A = -\frac{1}{4}, \quad B = \frac{1}{4}, \quad C = 0, \quad D = -\frac{1}{2}
$$
So:
$$
\frac{1}{x^4-1} = -\frac{1}{4(x+1)} + \frac{1}{4(x-1)} – \frac{1}{2(x^2+1)}
$$
Step 1: Integrate each term
$$
\int \frac{1}{x^4-1} \; dx = \int -\frac{1}{4(x+1)} \; dx + \int \frac{1}{4(x-1)}\; dx – \int \frac{1}{2(x^2+1)} \; dx
$$
$$
\int \frac{1}{x^4-1} \; dx = -\frac{1}{4} \ln|x+1| + \frac{1}{4} \ln|x-1| – \frac{1}{2} \tan^{-1}x + C
$$
$$
\int \frac{1}{x^4-1} \; dx = \frac{1}{4} \ln\left|\frac{x-1}{x+1}\right| – \frac{1}{2} \tan^{-1}x + C
$$
Final Answer
$$
\boxed{\frac{1}{4} \ln\left|\frac{x-1}{x+1}\right| – \frac{1}{2} \tan^{-1}x + C}
$$
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NCERT Question 16: Evaluate the integral
$$\displaystyle \int \frac{dx}{x\bigl(x^{n}+1\bigr)}$$
Solution
$$\displaystyle \int \frac{dx}{x\bigl(x^{n}+1\bigr)}$$
Let $t = x^{n}$ so $dt = n\; x^{,n-1}\;dx$ hence $x^{,n-1}\;dx = \dfrac{dt}{n}$
Rewriting the integrand by multiplying numerator and denominator by $x^{,n-1}$ gives
$$\displaystyle \int \frac{dx}{x\bigl(x^{n}+1\bigr)}=\int\frac{x^{,n-1}}{x^{n}\bigl(x^{n}+1\bigr)}dx =\int\frac{1}{x^{n}}\cdot\frac{x^{,n-1}}{x^{n}+1}dx $$
Substituting $t=x^{n}$ yields
$$\displaystyle \int \frac{dx}{x\bigl(x^{n}+1\bigr)}=\frac{1}{n}\int \frac{dt}{t(t+1)}$$
Partial fractions:
$$\displaystyle \frac{1}{t(t+1)}=\frac{1}{t}-\frac{1}{t+1}$$
So
$$\displaystyle \frac{1}{n}\int \frac{dt}{t(t+1)}=\frac{1}{n}\int\Big(\frac{1}{t}-\frac{1}{t+1}\Big)\;dt
=\frac{1}{n}\big(\ln|t|-\ln|t+1|\big)+C$$
Substitute back $t=x^{n}$:
$$\displaystyle \frac{1}{n}\int \frac{dt}{t(t+1)} =\frac{1}{n}\ln\left|\frac{x^{n}}{x^{n}+1}\right|+C$$
$$\boxed{\displaystyle \frac{1}{n}\ln\left|\frac{x^{n}}{x^{n}+1}\right|+C}$$
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NCERT Question 17: Evaluate the integral
$$\int \frac{\cos x}{(1-\sin x)(2-\sin x)}\ dx$$
Solution
$$\int \frac{\cos x}{(1-\sin x)(2-\sin x)}\ dx$$
Let $ \sin x = t$ so $ \cos x\ dx = dt$
$$\int \frac{\cos x}{(1-\sin x)(2-\sin x)}\ dx = \int \frac{dt}{(1-t)(2-t)}$$
Write
$$\frac{1}{(1-t)(2-t)}=\frac{A}{1-t}+\frac{B}{2-t}$$
so
$$1=A(2-t)+B(1-t)$$
Giving $A=1$ and $B=-1$
$$\int \frac{dt}{(1-t)(2-t)}=\int\left(\frac{1}{1-t}-\frac{1}{2-t}\right)\;dt = -\ln|1-t|+\ln|2-t|+C$$
Hence,
$$\int \frac{dt}{(1-t)(2-t)} = \ln\left|\frac{2-t}{1-t}\right| + C $$
Substitute back $t=\sin x$:
$$\int \frac{\cos x}{(1-\sin x)(2-\sin x)}\ dx= \ln\left|\frac{2-\sin x}{1-\sin x}\right| + C$$
Final Answer :
$$\boxed{\ln\left|\frac{2-\sin x}{1-\sin x}\right| + C}$$
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NCERT Question 18 : Evaluate the Integral
$$
\int \frac{(x^{2}+1)(x^{2}+2)}{(x^{2}+3)(x^{2}+4)} \; dx
$$
Solution
$$
\int \frac{(x^{2}+1)(x^{2}+2)}{(x^{2}+3)(x^{2}+4)} \; dx
$$
First expand the numerator:
$$
(x^{2}+1)(x^{2}+2)=x^{4}+3x^{2}+2
$$
Expand the denominator:
$$
(x^{2}+3)(x^{2}+4)=x^{4}+7x^{2}+12
$$
Convert into “1 – something” form
Because numerator and denominator have the same degree, try writing:
$$
\frac{x^{4}+3x^{2}+2}{x^{4}+7x^{2}+12}=1-\frac{\text{something}}{(x^{2}+3)(x^{2}+4)}
$$
Compute the “something”:
$$
(x^{4}+7x^{2}+12)-(x^{4}+3x^{2}+2)=4x^{2}+10
$$
Hence:
$$
\frac{(x^{2}+1)(x^{2}+2)}{(x^{2}+3)(x^{2}+4)}
=1-\frac{4x^{2}+10}{(x^{2}+3)(x^{2}+4)}
$$
Write the rational expression as:
$$
\frac{x^{4}+3x^{2}+2}{x^{4}+7x^{2}+12}
=1-\frac{4x^{2}+10}{(x^{2}+3)(x^{2}+4)}
$$
Now decompose:
$$
\frac{4x^{2}+10}{(x^{2}+3)(x^{2}+4)}
=\frac{Ax+B}{x^{2}+3}+\frac{Cx+D}{x^{2}+4}
$$
Multiply both sides by $(x^{2}+3)(x^{2}+4)$:
$$
4x^{2}+10=(Ax+B)(x^{2}+4)+(Cx+D)(x^{2}+3)
$$
Expand:
$$
4x^{2}+10=(A+C)x^{3}+(B+D)x^{2}+(4A+3C)x+(4B+3D)
$$
Equating coefficients:
$$
A+C=0
$$
$$
B+D=4
$$
$$
4A+3C=0
$$
$$
4B+3D=10
$$
Solving:
$$
A=0,; B=-2,; C=0,; D=6
$$
Thus:
$$
\frac{4x^{2}+10}{(x^{2}+3)(x^{2}+4)}
=-\frac{2}{x^{2}+3}+\frac{6}{x^{2}+4}
$$
So the integrand becomes:
$$1-\frac{4x^{2}+10}{(x^{2}+3)(x^{2}+4)} = 1+\frac{2}{x^{2}+3}-\frac{6}{x^{2}+4}$$
Now integrate term-by-term:
$$
\int 1 \; dx = x
$$
$$
\int \frac{2}{x^{2}+3} \; dx
= \frac{2}{\sqrt{3}} \tan^{-1}\left( \frac{x}{\sqrt{3}} \right)
$$
$$
\int \frac{6}{x^{2}+4} \; dx
= 3 \tan^{-1}\left( \frac{x}{2} \right)
$$
Hence,
$$ \int \frac{(x^{2}+1)(x^{2}+2)}{(x^{2}+3)(x^{2}+4)} \; dx = x + \frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{x}{\sqrt{3}}\right)-3 \tan^{-1}\left(\frac{x}{2}\right) + C $$
Final Answer
$$\boxed{x + \frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{x}{\sqrt{3}}\right)-3 \tan^{-1}\left(\frac{x}{2}\right) + C}$$
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NCERT Question 19: Evaluate the integral
$$\int \frac{2x}{(x^{2}+1)(x^{2}+3)}\ dx $$
Solution
$$\int \frac{2x}{(x^{2}+1)(x^{2}+3)}\ dx $$
Let $t = x^{2}$ so $dt = 2x\ dx$.
$$
\int \frac{2x}{(x^{2}+1)(x^{2}+3)}\ dx
= \int \frac{dt}{(t+1)(t+3)}
$$
Decompose into partial fractions:
$$
\frac{1}{(t+1)(t+3)}=\frac{A}{t+1}+\frac{B}{t+3}
$$
Solving $1=A(t+3)+B(t+1)$ gives $A=\tfrac{1}{2}$ and $B=-\tfrac{1}{2}$. Hence
$$
\int \frac{dt}{(t+1)(t+3)}
= \int\left(\frac{1}{2(t+1)}-\frac{1}{2(t+3)}\right)dt
$$
$$
\int\left(\frac{1}{2(t+1)}-\frac{1}{2(t+3)}\right)dt = \frac{1}{2}\ln|t+1|-\frac{1}{2}\ln|t+3|+C
$$
$$
\int \frac{dt}{(t+1)(t+3)}
= \frac{1}{2}\ln\left|\frac{t+1}{t+3}\right|+C
$$
Substitute back $t=x^{2}$:
$$\int \frac{2x}{(x^{2}+1)(x^{2}+3)}\ dx = \frac{1}{2}\ln\left|\frac{x^{2}+1}{x^{2}+3}\right|+C $$
Final Answer
$$
\boxed{\frac{1}{2}\ln\left|\frac{x^{2}+1}{x^{2}+3}\right|+C}
$$
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NCERT Question 20: Evaluate the integral
$$ \int \frac{1}{x\bigl(x^{4}-1\bigr)}\ dx $$
Solution
$$ \int \frac{1}{x\bigl(x^{4}-1\bigr)}\ dx $$
Multiply numerator and denominator by $x^{3}$ and substitute $t=x^{4}$:
$$
\int \frac{1}{x\bigl(x^{4}-1\bigr)}dx=\int \frac{x^{3}}{x^{4}\bigl(x^{4}-1\bigr)}dx
$$
Let $t=x^{4}$ so $dt=4x^{3}\;dx$ hence $x^{3}\;dx=\dfrac{dt}{4}$. Therefore
$$
\int \frac{1}{x\bigl(x^{4}-1\bigr)}\ dx
= \int \frac{x^{3}}{x^{4}\bigl(x^{4}-1\bigr)}\ dx
= \frac{1}{4}\int \frac{dt}{t(t-1)}
$$
Decompose the integrand:
$$
\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{t-1}
$$
Solving $1=A(t-1)+Bt$ gives $A=-1,;B=1$, so
$$
\frac{1}{t(t-1)}=-\frac{1}{t}+\frac{1}{t-1}.
$$
Integrate:
$$
\frac{1}{4}\int\left(-\frac{1}{t}+\frac{1}{t-1}\right)dt
=\frac{1}{4}\big(-\ln|t|+\ln|t-1|\big)+C
=\frac{1}{4}\ln\left|\frac{t-1}{t}\right|+C
$$
Substitute back $t=x^{4}$:
$$ \int \frac{1}{x\bigl(x^{4}-1\bigr)}\ dx = \frac{1}{4}\ln\left|\frac{x^{4}-1}{x^{4}}\right|+C $$
Final Answer
$$
\boxed{\displaystyle \frac{1}{4}\ln\left|\frac{x^{4}-1}{x^{4}}\right|+C}
$$
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NCERT Question 21: Evaluate the integral
$$\int \frac{1}{e^{x}-1}\ dx$$
Solution
$$\int \frac{1}{e^{x}-1}\ dx$$
Let $t = e^{x}$ so $dt = e^{x}\;dx = t\;dx$ hence $dx = \dfrac{dt}{t}$.
$$\int \frac{1}{e^{x}-1}\ dx = \int \frac{1}{t-1}\cdot\frac{dt}{t} = \int \frac{1}{t(t-1)}\ dt$$
Decompose into partial fractions:
$$\frac{1}{t(t-1)} = -\frac{1}{t} + \frac{1}{t-1}$$
Integrate:
$$\int\left(-\frac{1}{t}+\frac{1}{t-1}\right)\;dt
= -\ln|t| + \ln|t-1| + C
= \ln\left|\frac{t-1}{t}\right| + C$$
Substitute back $t=e^{x}$:
$$\int \frac{1}{e^{x}-1}\ dx = \ln\left|\frac{e^{x}-1}{e^{x}}\right| + C$$
Final Answer
$$\boxed{\ln\left|\frac{e^{x}-1}{e^{x}}\right| + C}
\quad\text{(equivalently }\boxed{\ln!\big|1-e^{-x}\big| + C}\text{)}$$
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NCERT Question 22: Choose the correct answer by evaluate the integral
$$\int \frac{x}{(x-1)(x-2)}\ dx$$
Options
(A) $\ln\left|\dfrac{(x-1)^{2}}{x-2}\right| + C$
(B) $\ln\left|\dfrac{(x-2)^{2}}{x-1}\right| + C$
(C) $\ln\left|\dfrac{x-1}{x-2}\right|^{2} + C$
(D) $\ln\left|(x-1)(x-2)\right| + C$
Solution
$$\int \frac{x}{(x-1)(x-2)}\ dx$$
Decompose into partial fractions:
$$\frac{x}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}$$
Multiply through:
$$x=A(x-2)+B(x-1)=(A+B)x+(-2A-B)$$
Equate coefficients:
$$A+B=1$$
$$-2A-B=0$$
Solve the system:
From $-2A-B=0$ we get $B=-2A$. Substitute into $A+B=1$:
$$A-2A=1\Rightarrow A=-1$$
$$B=-2A=2$$
So
$$\frac{x}{(x-1)(x-2)}=-\frac{1}{x-1}+\frac{2}{x-2}$$
Integrate termwise:
$$\int \frac{x}{(x-1)(x-2)}\ dx
=-\int \frac{1}{x-1}\ dx + 2\int \frac{1}{x-2}\ dx$$
$$-\int \frac{1}{x-1}\ dx + 2\int \frac{1}{x-2}\ dx =-\ln|x-1| + 2\ln|x-2| + C $$
$$ \int \frac{x}{(x-1)(x-2)}\ dx=\ln\left|\frac{(x-2)^{2}}{x-1}\right| + C$$
Correct option: (B)
$$\boxed{\ln\left|\dfrac{(x-2)^{2}}{x-1}\right| + C}$$
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NCERT Question 23: Choose the correct answer by evaluate the integral
$$\int \frac{dx}{x(x^{2}+1)}$$
Options
(A) $\displaystyle \ln|x|-\tfrac{1}{2}\ln!\big(x^{2}+1\big)+C$
(B) $\displaystyle \ln|x|+\tfrac{1}{2}\ln!\big(x^{2}+1\big)+C$
(C) $\displaystyle -\ln|x|+\tfrac{1}{2}\ln!\big(x^{2}+1\big)+C$
(D) $\displaystyle \tfrac{1}{2}\ln|x|+\ln!\big(x^{2}+1\big)+C$
Solution
$$\int \frac{dx}{x(x^{2}+1)}$$
Decompose into partial fractions:
$$
\frac{1}{x(x^{2}+1)}=\frac{A}{x}+\frac{Bx+C}{x^{2}+1}
$$
Multiply both sides by $x(x^{2}+1)$ and compare coefficients:
$$
1 = A(x^{2}+1)+(Bx+C)x = (A+B)x^{2}+Cx+A
$$
Equating coefficients gives
$$
A+B=0,\quad C=0,\quad A=1
$$
So
$$
A=1, \quad B=-1, \quad C=0
$$
Thus
$$
\frac{1}{x(x^{2}+1)}=\frac{1}{x}-\frac{x}{x^{2}+1}
$$
Integrate termwise:
$$
\int \frac{dx}{x(x^{2}+1)}=\int \frac{1}{x}\;dx \;-\; \int \frac{x}{x^{2}+1}\;dx
$$
For the second integral let $u=x^{2}+1$ so $du=2x\;dx$:
$$
\int \frac{x}{x^{2}+1}\;dx=\frac{1}{2}\ln\big(x^{2}+1\big)
$$
Therefore
$$
\int \frac{dx}{x(x^{2}+1)}=\ln|x|-\frac{1}{2}\ln\big(x^{2}+1\big)+C
$$
Correct option: (A)
$$
\boxed{\ln|x|-\dfrac{1}{2}\ln\big(x^{2}+1\big)+C}
$$
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