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NCERT Question 12: Find the Integral
$$\int \frac{x^3 + 3x + 4}{\sqrt{x}}dx$$
Solution:
$$\int \frac{x^3 + 3x + 4}{\sqrt{x}}dx$$
Simplify the integrand:
$$
\frac{x^3 + 3x + 4}{\sqrt{x}} = x^{\frac{5}{2}} + 3x^{\frac{1}{2}} + 4x^{-\frac{1}{2}}
$$
Therefore,
$$
\int \frac{x^3 + 3x + 4}{\sqrt{x}}dx = \int \left(x^{\frac{5}{2}} + 3x^{\frac{1}{2}} + 4x^{-\frac{1}{2}}\right)dx
$$
Integrating each term,
$$
= \frac{x^{\frac{7}{2}}}{\frac{7}{2}} + 3\frac{x^{\frac{3}{2}}}{\frac{3}{2}} + 4\frac{x^{\frac{1}{2}}}{\frac{1}{2}} + C
$$
Simplify the fractions:
$$
= \frac{2}{7}x^{\frac{7}{2}} + 2x^{\frac{3}{2}} + 8x^{\frac{1}{2}} + C
$$
or equivalently,
$$
\boxed{\frac{2}{7}x^{\frac{7}{2}} + 2x^{\frac{3}{2}} + 8\sqrt{x} + C}
$$
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NCERT Question 13: Find the Integral
$$\int \frac{x^3 – x^2 + x – 1}{x – 1}dx$$
Solution:
$$\int \frac{x^3 – x^2 + x – 1}{x – 1}dx$$
Divide the numerator by the denominator:
$$
\frac{x^3 – x^2 + x – 1}{x – 1} = x^2 + 1
$$
Therefore,
$$
\int \frac{x^3 – x^2 + x – 1}{x – 1}dx = \int (x^2 + 1)dx
$$
Integrating each term,
$$
= \int x^2dx + \int 1dx
$$
$$
= \frac{x^3}{3} + x + C
$$
Hence,
$$
\boxed{\frac{x^3}{3} + x + C}
$$
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NCERT Question 14: Find the Integral
$$\int (1 – x)\sqrt{x}dx$$
Solution:
$$\int (1 – x)\sqrt{x}dx$$
Simplify the integrand:
$$
(1 – x)\sqrt{x} = \sqrt{x} – x\sqrt{x} = x^{\frac{1}{2}} – x^{\frac{3}{2}}
$$
Therefore,
$$
\int (1 – x)\sqrt{x}dx = \int x^{\frac{1}{2}}dx – \int x^{\frac{3}{2}}dx
$$
Integrating each term,
$$
= \frac{x^{\frac{3}{2}}}{\frac{3}{2}} – \frac{x^{\frac{5}{2}}}{\frac{5}{2}} + C
$$
Simplify the fractions:
$$
= \frac{2}{3}x^{\frac{3}{2}} – \frac{2}{5}x^{\frac{5}{2}} + C
$$
Hence,
$$
\boxed{\frac{2}{3}x^{\frac{3}{2}} – \frac{2}{5}x^{\frac{5}{2}} + C}
$$
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NCERT Question 15: Find the Integral
$$\int \sqrt{x}(3x^2 + 2x + 3)dx$$
Solution:
$$\int \sqrt{x}(3x^2 + 2x + 3)dx$$
Simplify the integrand:
$$
\sqrt{x}(3x^2 + 2x + 3) = 3x^{\frac{5}{2}} + 2x^{\frac{3}{2}} + 3x^{\frac{1}{2}}
$$
Therefore,
$$
\int \sqrt{x}(3x^2 + 2x + 3)dx = 3\int x^{\frac{5}{2}}dx + 2\int x^{\frac{3}{2}}dx + 3\int x^{\frac{1}{2}}dx
$$
Integrating each term,
$$
= 3\frac{x^{\frac{7}{2}}}{\frac{7}{2}} + 2\frac{x^{\frac{5}{2}}}{\frac{5}{2}} + 3\frac{x^{\frac{3}{2}}}{\frac{3}{2}} + C
$$
Simplify the coefficients:
$$
= \frac{6x^{\frac{7}{2}}}{7} + \frac{4x^{\frac{5}{2}}}{5} + 2x^{\frac{3}{2}} + C
$$
Hence,
$$
\boxed{\frac{6x^{\frac{7}{2}}}{7} + \frac{4x^{\frac{5}{2}}}{5} + 2x^{\frac{3}{2}} + C}
$$
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NCERT Question 16: Find the Integral
$$\int (2x – 3\cos x + e^x)dx$$
Solution:
$$
\int (2x – 3\cos x + e^x)dx = 2\int xdx – 3\int \cos xdx + \int e^x dx
$$
Integrating each term:
$$
= 2\left(\frac{x^2}{2}\right) – 3(\sin x) + e^x + C
$$
Simplify:
$$
= x^2 – 3\sin x + e^x + C
$$
Hence,
$$
\boxed{x^2 – 3\sin x + e^x + C}
$$
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NCERT Question 17: Find the Integral
$$\int (2x^2 – 3\sin x + 5\sqrt{x})dx$$
Solution:
$$
\int (2x^2 – 3\sin x + 5\sqrt{x})dx = 2\int x^2dx – 3\int \sin xdx + 5\int \sqrt{x}dx
$$
Integrating each term:
$$
= 2\left(\frac{x^3}{3}\right) – 3(-\cos x) + 5\left(\frac{x^{3/2}}{3/2}\right) + C
$$
Simplify:
$$
= \frac{2x^3}{3} + 3\cos x + \frac{10x^{3/2}}{3} + C
$$
Hence,
$$
\boxed{\frac{2x^3}{3} + 3\cos x + \frac{10x^{3/2}}{3} + C}
$$
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NCERT Question 18: Find the Integral
$$\int \sec x(\sec x + \tan x)dx$$
Solution:
$$
\int \sec x(\sec x + \tan x)dx = \int (\sec^2 x + \sec x \tan x)dx
$$
$$
= \int \sec^2 xdx + \int \sec x \tan xdx
$$
$$
= \tan x + \sec x + C
$$
Hence,
$$
\boxed{\tan x + \sec x + C}
$$
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NCERT Question 19: Find the Integral
$$\int \frac{\sec^2 x}{\csc^2 x}dx$$
Solution:
$$
\int \frac{\sec^2 x}{\csc^2 x}dx = \int \frac{1/\cos^2 x}{1/\sin^2 x}dx
$$
$$
= \int \frac{\sin^2 x}{\cos^2 x}dx = \int \tan^2 xdx
$$
Now, using the identity $\tan^2 x = \sec^2 x – 1$
$$
\int \tan^2 xdx = \int (\sec^2 x – 1)dx
$$
$$
= \int \sec^2 xdx – \int 1dx
$$
$$
= \tan x – x + C
$$
Hence,
$$
\boxed{\tan x – x + C}
$$
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NCERT Question 20 : Find the Integral
$$\int \frac{2 – 3\sin x}{\cos^2 x}dx$$
Solution:
$$
\int \frac{2 – 3\sin x}{\cos^2 x}dx = \int \left(\frac{2}{\cos^2 x} – \frac{3\sin x}{\cos^2 x}\right)dx
$$
$$
= 2\int \sec^2 xdx – 3\int \tan x \sec xdx
$$
$$
= 2\tan x – 3\sec x + C
$$
Hence,
$$
\boxed{2\tan x – 3\sec x + C}
$$
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NCERT Question 21 : The anti-derivative of $(\sqrt{x} + \frac{1}{\sqrt{x}})$ equals
(A) $\frac{1}{3}x^{1/3} + 2x^{1/2} + C$
(B) $\frac{2}{3}x^{2/3} + \frac{1}{2}x^2 + C$
(C) $\frac{2}{3}x^{3/2} + 2x^{1/2} + C$
(D) $\frac{3}{2}x^{3/2} + \frac{1}{2}x^{1/2} + C$
Solution:
$$
\int \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)dx = \int \left(x^{1/2} + x^{-1/2}\right)dx
$$
$$
= \int x^{1/2}dx + \int x^{-1/2}dx
$$
$$
= \frac{x^{3/2}}{3/2} + \frac{x^{1/2}}{1/2} + C
$$
$$
= \frac{2}{3}x^{3/2} + 2x^{1/2} + C
$$
Hence, the correct option is
$$
\boxed{(C);\frac{2}{3}x^{3/2} + 2x^{1/2} + C}
$$
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NCERT Question 22 : If $\frac{d}{dx}f(x) = 4x^3 – \frac{3}{x^4}$ such that $f(2) = 0$, then $f(x)$ is
(A) $x^4 + \frac{1}{x^3} – \frac{129}{8}$
(B) $x^3 + \frac{1}{x^4} + \frac{129}{8}$
(C) $x^4 + \frac{1}{x^3} + \frac{129}{8}$
(D) $x^3 + \frac{1}{x^4} – \frac{129}{8}$
Solution:
$$
f(x) = \int \left(4x^3 – \frac{3}{x^4}\right)dx
$$
$$
= 4\int x^3dx – 3\int x^{-4}dx
$$
$$
= 4\left(\frac{x^4}{4}\right) – 3\left(\frac{x^{-3}}{-3}\right) + C
$$
$$
= x^4 + \frac{1}{x^3} + C
$$
Now using $f(2) = 0$:
$$
f(2) = (2)^4 + \frac{1}{(2)^3} + C = 0
$$
$$
16 + \frac{1}{8} + C = 0
$$
$$
C = -\frac{129}{8}
$$
Therefore,
$$
f(x) = x^4 + \frac{1}{x^3} – \frac{129}{8}
$$
Hence, the correct option is
$$
\boxed{(A);x^4 + \frac{1}{x^3} – \frac{129}{8}}
$$
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