Solution

Let $1+x^{2}=t$

$\therefore 2 x d x=d t$

$\Rightarrow \int \frac{2 x}{1+x^{2}} d x=\int \frac{1}{t} d t$

$=\log |t|+C$

$=\log |1+x^{2}|+C$

$=\log (1+x^{2})+C$

Solution

Let $\log |x|=t$

$\therefore \frac{1}{x} d x=d t$

$ \begin{aligned} \Rightarrow \int \frac{(\log |x|)^{2}}{x} d x & =\int t^{2} d t \\ & =\frac{t^{3}}{3}+C \\ & =\frac{(\log |x|)^{3}}{3}+C \end{aligned} $

Solution

$\frac{1}{x+x \log x}=\frac{1}{x(1+\log x)}$

Let $1+\log x=t$

$\therefore \frac{1}{x} d x=d t$

$\Rightarrow \int \frac{1}{x(1+\log x)} d x=\int \frac{1}{t} d t$

$=\log |t|+C$

$=\log |1+\log x|+C$

Solution

$\sin x \cdot \sin (\cos x)$

Let $\cos x=t$

$\therefore-\sin x d x=d t$

$ \begin{aligned} \Rightarrow \int \sin x \cdot \sin (\cos x) d x & =-\int \sin t d t \\ & =-[-\cos t]+C \\ & =\cos t+C \\ & =\cos (\cos x)+C \end{aligned} $

Solution

$\sin (a x+b) \cos (a x+b)=\frac{2 \sin (a x+b) \cos (a x+b)}{2}=\frac{\sin 2(a x+b)}{2}$

Let $2(a x+b)=t$

$\therefore 2 a d x=d t$ $\Rightarrow \int \frac{\sin 2(a x+b)}{2} d x=\frac{1}{2} \int \frac{\sin t d t}{2 a}$

$ \begin{aligned} & =\frac{1}{4 a}[-\cos t]+C \\ & =\frac{-1}{4 a} \cos 2(a x+b)+C \end{aligned} $

Solution

Let $a x+b=t$

$\Rightarrow a d x=d t$

$\therefore d x=\frac{1}{a} d t$

$\Rightarrow \int(a x+b)^{\frac{1}{2}} d x=\frac{1}{a} \int t^{\frac{1}{2}} d t$

$=\frac{1}{a}(\frac{t^{\frac{3}{2}}}{\frac{3}{2}})+C$

$=\frac{2}{3 a}(a x+b)^{\frac{3}{2}}+C$

Solution

Let $(x+2)=t$ $\therefore d x=d t$

$ \begin{aligned} \Rightarrow \int x \sqrt{x+2} d x & =\int(t-2) \sqrt{t} d t \\ & =\int(t^{\frac{3}{2}}-2 t^{\frac{1}{2}}) d t \\ & =\int t^{\frac{3}{2}} d t-2 \int t^{\frac{1}{2}} d t \\ & =\frac{t^{\frac{5}{2}}}{\frac{5}{2}}-2(\frac{t^{\frac{3}{2}}}{\frac{3}{2}})+C \\ & =\frac{2}{5} t^{\frac{5}{2}}-\frac{4}{3} t^{\frac{3}{2}}+C \\ & =\frac{2}{5}(x+2)^{\frac{5}{2}}-\frac{4}{3}(x+2)^{\frac{3}{2}}+C \end{aligned} $

Solution

Let $1+2 x^{2}=t$

$\therefore 4 x d x=d t$ $\Rightarrow \int x \sqrt{1+2 x^{2}} d x=\int \frac{\sqrt{t} d t}{4}$

$=\frac{1}{4} \int t^{\frac{1}{2}} d t$

$=\frac{1}{4}(\frac{t^{\frac{3}{2}}}{\frac{3}{2}})+C$

$=\frac{1}{6}(1+2 x^{2})^{\frac{3}{2}}+C$

Solution

Let $x^{2}+x+1=t$

$\therefore(2 x+1) d x=d t$

$\int(4 x+2) \sqrt{x^{2}+x+1} d x$

$=\int 2 \sqrt{t} d t$

$=2 \int \sqrt{t} d t$

$=2(\frac{t^{\frac{3}{2}}}{\frac{3}{2}})+C$

$=\frac{4}{3}(x^{2}+x+1)^{\frac{3}{2}}+C$

Solution

$\frac{1}{x-\sqrt{x}}=\frac{1}{\sqrt{x}(\sqrt{x}-1)}$

Let $(\sqrt{x}-1)=t$

$\therefore \frac{1}{2 \sqrt{x}} d x=d t$

$\Rightarrow \int \frac{1}{\sqrt{x}(\sqrt{x}-1)} d x=\int \frac{2}{t} d t$

$=2 \log |t|+C$

$=2 \log |\sqrt{x}-1|+C$

Solution

Let $x+4=t$

$\therefore d x=d t$

$ \begin{aligned} \int \frac{x}{\sqrt{x+4}} d x & =\int \frac{(t-4)}{\sqrt{t}} d t \\ & =\int(\sqrt{t}-\frac{4}{\sqrt{t}}) d t \\ & =\frac{t^{\frac{3}{2}}}{\frac{3}{2}}-4(\frac{t^{\frac{1}{2}}}{\frac{1}{2}})+C \\ & =\frac{2}{3}(t)^{\frac{3}{2}}-8(t)^{\frac{1}{2}}+C \\ & =\frac{2}{3} t \cdot t^{\frac{1}{2}}-8 t^{\frac{1}{2}}+C \\ & =\frac{2}{3} t^{\frac{1}{2}}(t-12)+C \\ & =\frac{2}{3}(x+4)^{\frac{1}{2}}(x+4-12)+C \\ & =\frac{2}{3} \sqrt{x+4}(x-8)+C \end{aligned} $

Solution

Let $x^{3}-1=t$

$\therefore 3 x^{2} d x=d t$ $\Rightarrow \int(x^{3}-1)^{\frac{1}{3}} x^{5} d x=\int(x^{3}-1)^{\frac{1}{3}} x^{3} \cdot x^{2} d x$

$=\int t^{\frac{1}{3}}(t+1) \frac{d t}{3}$

$=\frac{1}{3} \int(t^{\frac{4}{3}}+t^{\frac{1}{3}}) d t$

$=\frac{1}{3}[\frac{t^{\frac{7}{3}}}{\frac{7}{3}}+\frac{t^{\frac{4}{3}}}{\frac{4}{3}}]+C$

$=\frac{1}{3}[\frac{3}{7} t^{\frac{7}{3}}+\frac{3}{4} t^{\frac{4}{3}}]+C$

$=\frac{1}{7}(x^{3}-1)^{\frac{7}{3}}+\frac{1}{4}(x^{3}-1)^{\frac{4}{3}}+C$

Solution

Let $2+3 x^{3}=t$

$\therefore 9 x^{2} d x=d t$

$ \begin{aligned} \Rightarrow \int \frac{x^{2}}{(2+3 x^{3})^{3}} d x & =\frac{1}{9} \int \frac{d t}{(t)^{3}} \\ & =\frac{1}{9}[\frac{t^{-2}}{-2}]+C \\ & =\frac{-1}{18}(\frac{1}{t^{2}})+C \\ & =\frac{-1}{18(2+3 x^{3})^{2}}+C \end{aligned} $

Solution

Let $\log x=t$

$\therefore \frac{1}{x} d x=d t$

$\Rightarrow \int \frac{1}{x(\log x)^{m}} d x=\int \frac{d t}{(t)^{m}}$

$=(\frac{t^{-m+1}}{1-m})+C$

$=\frac{(\log x)^{1-m}}{(1-m)}+C$

Solution

Let $9-4 x^{2}=t$

$\therefore-8 x d x=d t$

$ \begin{aligned} \Rightarrow \int \frac{x}{9-4 x^{2}} d x & =\frac{-1}{8} \int_t^{1} d t \\ & =\frac{-1}{8} \log |t|+C \\ & =\frac{-1}{8} \log |9-4 x^{2}|+C \end{aligned} $

Solution

Let $2 x+3=t$

$\therefore 2 d x=d t$

$ \begin{aligned} \Rightarrow \int e^{2 x+3} d x & =\frac{1}{2} \int e^{t} d t \\ & =\frac{1}{2}(e^{t})+C \\ & =\frac{1}{2} e^{(2 x+3)}+C \end{aligned} $

Solution

Let $x^{2}=t$

$\therefore 2 x d x=d t$

$\Rightarrow \int \frac{x}{e^{x^{2}}} d x=\frac{1}{2} \int \frac{1}{e^{t}} d t$

$=\frac{1}{2} \int e^{-t} d t$

$=\frac{1}{2}(\frac{e^{-t}}{-1})+C$

$=-\frac{1}{2} e^{-x^{2}}+C$

$=\frac{-1}{2 e^{x^{2}}}+C$

Solution

Let $\tan ^{-1} x=t$

$\therefore \frac{1}{1+x^{2}} d x=d t$ $\Rightarrow \int \frac{e^{\tan ^{-1} x}}{1+x^{2}} d x=\int e^{t} d t$

$=e^{t}+C$

$=e^{\tan ^{-1} x}+C$

Solution

$\frac{e^{2 x}-1}{e^{2 x}+1}$

Dividing numerator and denominator by $e^{x}$, we obtain

$ \frac{\frac{(e^{2 x}-1)}{e^{x}}}{(e^{2 x}+1)}=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} $

Let $e^{x}+e^{-x}=t$

$\therefore(e^{x}-e^{-x}) d x=d t$

$\Rightarrow \int \frac{e^{2 x}-1}{e^{2 x}+1} d x=\int \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} d x$

$=\int \frac{d t}{t}$

$=\log |t|+C$

$=\log |e^{x}+e^{-x}|+C$

Solution

Let $e^{2 x}+e^{-2 x}=t$

$\therefore(2 e^{2 x}-2 e^{-2 x}) d x=d t$

$\Rightarrow 2(e^{2 x}-e^{-2 x}) d x=d t$

$\Rightarrow \int(\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}) d x=\int \frac{d t}{2 t}$

$=\frac{1}{2} \int_t^{1} d t$

$=\frac{1}{2} \log |t|+C$

$=\frac{1}{2} \log |e^{2 x}+e^{-2 x}|+C$

Solution

$\tan ^{2}(2 x-3)=\sec ^{2}(2 x-3)-1$

Let $2 x-3=t$

$\therefore 2 d x=d t$

$ \begin{aligned} \Rightarrow \int \tan ^{2}(2 x-3) d x & =\int[(\sec ^{2}(2 x-3))-1] d x \\ & =\frac{1}{2} \int(\sec ^{2} t) d t-\int 1 d x \\ & =\frac{1}{2} \int \sec ^{2} t d t-\int 1 d x \\ & =\frac{1}{2} \tan t-x+C \\ & =\frac{1}{2} \tan (2 x-3)-x+C \end{aligned} $

Solution

Let $7-4 x=t$

$\therefore-4 d x=d t$

$ \begin{aligned} \therefore \int \sec ^{2}(7-4 x) d x & =\frac{-1}{4} \int \sec ^{2} t d t \\ & =\frac{-1}{4}(\tan t)+C \\ & =\frac{-1}{4} \tan (7-4 x)+C \end{aligned} $

Solution

Let $\sin ^{-1} x=t$ $\therefore \frac{1}{\sqrt{1-x^{2}}} d x=d t$

$\Rightarrow \int \frac{\sin ^{-1} x}{\sqrt{1-x^{2}}} d x=\int t d t$

$=\frac{t^{2}}{2}+C$

$=\frac{(\sin ^{-1} x)^{2}}{2}+C$

Solution

$\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}=\frac{2 \cos x-3 \sin x}{2(3 \cos x+2 \sin x)}$

Let $3 \cos x+2 \sin x=t$

$\therefore(-3 \sin x+2 \cos x) d x=d t$

$ \begin{aligned} \int \frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x} d x & =\int \frac{d t}{2 t} \\ & =\frac{1}{2} \int \frac{1}{t} d t \\ & =\frac{1}{2} \log |t|+C \\ & =\frac{1}{2} \log |2 \sin x+3 \cos x|+C \end{aligned} $

Solution

$\frac{1}{\cos ^{2} x(1-\tan x)^{2}}=\frac{\sec ^{2} x}{(1-\tan x)^{2}}$

Let $(1-\tan x)=t$

$\therefore-\sec ^{2} x d x=d t$

$ \begin{aligned} \Rightarrow \int \frac{\sec ^{2} x}{(1-\tan x)^{2}} d x & =\int \frac{-d t}{t^{2}} \\ & =-\int t^{-2} d t \\ & =+\frac{1}{t}+C \\ & =\frac{1}{(1-\tan x)}+C \end{aligned} $

Solution

Let $\sqrt{x}=t$

$\therefore \frac{1}{2 \sqrt{x}} d x=d t$

$ \begin{aligned} \Rightarrow \int \frac{\cos \sqrt{x}}{\sqrt{x}} d x & =2 \int \cos t d t \\ & =2 \sin t+C \\ & =2 \sin \sqrt{x}+C \end{aligned} $

Solution

Let $\sin 2 x=t$

$\therefore 2 \cos 2 x d x=d t$

$ \begin{aligned} \Rightarrow \int \sqrt{\sin 2 x} \cos 2 x d x & =\frac{1}{2} \int \sqrt{t} d t \\ & =\frac{1}{2}(\frac{t^{\frac{3}{2}}}{\frac{3}{2}})+C \\ & =\frac{1}{3} t^{\frac{3}{2}}+C \\ & =\frac{1}{3}(\sin 2 x)^{\frac{3}{2}}+C \end{aligned} $

Solution

Let $1+\sin x=t$ $\therefore \cos x d x=d t$

$ \begin{aligned} \Rightarrow \int \frac{\cos x}{\sqrt{1+\sin x}} d x & =\int \frac{d t}{\sqrt{t}} \\ & =\frac{t^{\frac{1}{2}}}{\frac{1}{2}}+C \\ & =2 \sqrt{t}+C \\ & =2 \sqrt{1+\sin x}+C \end{aligned} $

Solution

Let $\log \sin x=t$

$\Rightarrow \frac{1}{\sin x} \cdot \cos x d x=d t$

$\therefore \cot x d x=d t$

$\Rightarrow \int \cot x \log \sin x d x=\int t d t$

$ \begin{aligned} & =\frac{t^{2}}{2}+C \\ & =\frac{1}{2}(\log \sin x)^{2}+C \end{aligned} $

Solution

Let $1+\cos x=t$ $\therefore-\sin x d x=d t$

$ \begin{aligned} \Rightarrow \int \frac{\sin x}{1+\cos x} d x & =\int-\frac{d t}{t} \\ & =-\log |t|+C \\ & =-\log |1+\cos x|+C \end{aligned} $

Solution

Let $1+\cos x=t$

$\therefore-\sin x d x=d t$

$ \begin{aligned} \Rightarrow \int \frac{\sin x}{(1+\cos x)^{2}} d x & =\int-\frac{d t}{t^{2}} \\ & =-\int t^{-2} d t \\ & =\frac{1}{t}+C \\ & =\frac{1}{1+\cos x}+\mathbf{C} \end{aligned} $

Solution

$ \text{ Let } \begin{aligned} I & =\int \frac{1}{1+\cot x} d x \\ & =\int \frac{1}{1+\frac{\cos x}{\sin x}} d x \\ & =\int \frac{\sin x}{\sin x+\cos x} d x \\ & =\frac{1}{2} \int \frac{2 \sin x}{\sin x+\cos x} d x \\ & =\frac{1}{2} \int \frac{(\sin x+\cos x)+(\sin x-\cos x)}{(\sin x+\cos x)} d x \\ & =\frac{1}{2} \int 1 d x+\frac{1}{2} \int \frac{\sin x-\cos x}{\sin x+\cos x} d x \\ & =\frac{1}{2}(x)+\frac{1}{2} \int \frac{\sin x-\cos x}{\sin x+\cos x} d x \end{aligned} $

Let $\sin x+\cos x=t \Rightarrow(\cos x-\sin x) d x=d t$

$ \begin{aligned} \therefore I & =\frac{x}{2}+\frac{1}{2} \int \frac{-(d t)}{t} \\ & =\frac{x}{2}-\frac{1}{2} \log |t|+C \\ & =\frac{x}{2}-\frac{1}{2} \log |\sin x+\cos x|+C \end{aligned} $

Solution

$ \text{ Let } \begin{aligned} I & =\int \frac{1}{1-\tan x} d x \\ & =\int \frac{1}{1-\frac{\sin x}{\cos x}} d x \\ & =\int \frac{\cos x}{\cos x-\sin x} d x \\ & =\frac{1}{2} \int \frac{2 \cos x}{\cos x-\sin x} d x \\ & =\frac{1}{2} \int \frac{(\cos x-\sin x)+(\cos x+\sin x)}{(\cos x-\sin x)} d x \\ & =\frac{1}{2} \int 1 d x+\frac{1}{2} \int \frac{\cos x+\sin x}{\cos x-\sin x} d x \\ & =\frac{x}{2}+\frac{1}{2} \int \frac{\cos x+\sin x}{\cos x-\sin x} d x \end{aligned} $

Put $\cos x-\sin x=t \Rightarrow(-\sin x-\cos x) d x=d t$

$ \begin{aligned} \therefore I & =\frac{x}{2}+\frac{1}{2} \int \frac{-(d t)}{t} \\ & =\frac{x}{2}-\frac{1}{2} \log |t|+C \\ & =\frac{x}{2}-\frac{1}{2} \log |\cos x-\sin x|+C \end{aligned} $

Solution

$ \text{ Let } \begin{aligned} I & =\int \frac{\sqrt{\tan x}}{\sin x \cos x} d x \\ & =\int \frac{\sqrt{\tan x} \times \cos x}{\sin x \cos x \times \cos x} d x \\ & =\int \frac{\sqrt{\tan x}}{\tan x \cos ^{2} x} d x \\ & =\int \frac{\sec ^{2} x d x}{\sqrt{\tan x}} \end{aligned} $

Let $\tan x=t \Rightarrow \sec ^{2} x d x=d t$

$ \begin{aligned} \therefore I & =\int \frac{d t}{\sqrt{t}} \\ & =2 \sqrt{t}+C \\ & =2 \sqrt{\tan x}+C \end{aligned} $

Solution

$\frac{(x+1)(x+\log x)^{2}}{x}=(\frac{x+1}{x})(x+\log x)^{2}=(1+\frac{1}{x})(x+\log x)^{2}$

Let $(x+\log x)=t$

$\therefore(1+\frac{1}{x}) d x=d t$

$\Rightarrow \int(1+\frac{1}{x})(x+\log x)^{2} d x=\int t^{2} d t$

$ \begin{aligned} & =\frac{t^{3}}{3}+C \\ & =\frac{1}{3}(x+\log x)^{3}+C \end{aligned} $

Solution

$\frac{(x+1)(x+\log x)^{2}}{x}=(\frac{x+1}{x})(x+\log x)^{2}=(1+\frac{1}{x})(x+\log x)^{2}$

Let $(x+\log x)=t$

$\therefore(1+\frac{1}{x}) d x=d t$

$\Rightarrow \int(1+\frac{1}{x})(x+\log x)^{2} d x=\int t^{2} d t$

$ \begin{aligned} & =\frac{t^{3}}{3}+C \\ & =\frac{1}{3}(x+\log x)^{3}+C \end{aligned} $

Solution

Let $x^{4}=t$

$\therefore 4 x^{3} d x=d t$ $\Rightarrow \int \frac{x^{3} \sin (\tan ^{-1} x^{4})}{1+x^{8}} d x=\frac{1}{4} \int \frac{\sin (\tan ^{-1} t)}{1+t^{2}} d t$

Let $\tan ^{-1} t=u$

$\therefore \frac{1}{1+t^{2}} d t=d u$

From (1), we obtain

$ \begin{aligned} & \begin{aligned} & \int \frac{x^{3} \sin (\tan ^{-1} x^{4}) d x}{1+x^{8}}=\frac{1}{4} \int \sin u d u \\ &=\frac{1}{4}(-\cos u)+C \\ &=\frac{-1}{4} \cos (\tan ^{-1} t)+C \\ &= \frac{-1}{4} \cos (\tan ^{-1} x^{4})+C \end{aligned} \end{aligned} $

Solution

Let $x^{10}+10^{x}=t$

$\therefore(10 x^{9}+10^{x} \log _{e} 10) d x=d t$ $\Rightarrow \int \frac{10 x^{9}+10^{x} \log _{e} 10}{x^{10}+10^{x}} d x=\int \frac{d t}{t}$

$=\log t+C$

$=\log (10^{x}+x^{10})+C$

Hence, the correct Answer is D.

Solution

Let $I=\int \frac{d x}{\sin ^{2} x \cos ^{2} x}$

$=\int \frac{1}{\sin ^{2} x \cos ^{2} x} d x$

$=\int \frac{\sin ^{2} x+\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x$

$=\int \frac{\sin ^{2} x}{\sin ^{2} x \cos ^{2} x} d x+\int \frac{\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x$

$=\int \sec ^{2} x d x+\int cosec^{2} x d x$

$=\tan x-\cot x+C$

Hence, the correct Answer is B.