Integrals

$\begin{array}{l}{\displaystyle \int x}logxdx=logx{\displaystyle \int x}dx-\int \frac{d}{dx}logx·{\displaystyle \int x}dxdx\\ =logx*\frac{x^2}{2}-\int \frac{1}{x}*\frac{x^2}{2}dx\\ =\frac{x^2}{2}·logx-\frac{1}{2}{\displaystyle \int x}dx\\ =\frac{x^2}{2}logx-\frac{1}{2}*\frac{x^2}{2}+c\\ =\frac{x^2}{2}logx-\frac{x^2}{4}+c.\end{array}$

$\begin{array}{l}\int x^2{e}^{x}dx=x^2\int e^{x}dx-\int \frac{d{x}^2}{dx}\int e^{x}dxdx\\ =x^2·e^x-{\displaystyle \int 2}x{e}^{x}dx\\ =x^2{e}^x-2[x\int e^{x}dx-\int \frac{dx}{dx}\int e^{x}dxdx]\\ =x^2{e}^x-2\left[x{e}^x-\int e^{x}dx\right]\\ =x^2{e}^x-2x{e}^x+2e^x+c\\ =e^x\left[x^2-2x+2\right]+c\end{array}$

$\begin{array}{l}{\displaystyle \int x}sin3xdx=x{\displaystyle \int sin}3xdx-\int \frac{dx}{dx}{\displaystyle \int sin}3xdxdx.\\ =-x\frac{cos3x}{3}+\int \frac{cos3x}{3}\\ =\frac{-xcos3x}{3}+\frac{sin3x}{3*3}+c\\ =-\frac{x}{3}cos3x+\frac{1}{9}sin3x+c\end{array}$

$\begin{array}{l}{\displaystyle \int x}sinxdx=x{\displaystyle \int sin}dx-{\displaystyle \int \frac{d}{dx}x{\displaystyle \int sinx}dxdx}\\ =x(-cosx)-{\displaystyle \int (}-cosx)dx\end{array}$

$\begin{array}{l}\frac{1}{x(x^2+1)}=\frac{x}{x^2(x^2+1)}\\ Putting,x^2=t\Rightarrow 2xdx=dt\end{array}$

$\begin{array}{l}{\displaystyle \int \frac{dx}{x(x^2+1)}}={\displaystyle \int \frac{xdx}{x^2(x^2+1)}=\frac{1}{2}}{\displaystyle \int \frac{dt}{t(t+1)}}\\ =\frac{1}{2}{\displaystyle \int \frac{(t+1)-t}{t(t+1)}}dt\\ =\frac{1}{2}\left\{{\displaystyle \int \frac{dt}{t}}-{\displaystyle \int \frac{dt}{t+1}}\right\}\\ =\frac{1}{2}\left[log\left|t\right|-log\left|t+1\right|\right]+\; C\\ =\frac{1}{2}log\left|x^2\right|-\frac{1}{2}log\left|x^2+1\right|+\; C\\ =\frac{2}{2}log\left|x\right|-\frac{1}{2}log\left|x^2+1\right|+c\\ =log\left|x\right|-\frac{1}{2}log\left|x^2+1\right|+c\\ So,option\left(A\right)iscorrect.\end{array}$

Kindly go through the solution

Putting e^x = t so that e^x dx = dt=>dx = $\frac{dt}{e^x}=\frac{dt}{t}$

$\begin{array}{l}\therefore {\displaystyle \int \frac{dx}{e^x-1}={\displaystyle \int \frac{dt}{(t-1)*t}={\displaystyle \int \frac{dt}{t(t-1)}}}}\\ ={\displaystyle \int \frac{t-(t-1)}{t(t-1)}}dt\\ ={\displaystyle \int \frac{dt}{t-1}}-{\displaystyle \int \frac{dt}{t}}\\ =log\left|t-1\right|-log\left|t\right|+\; C\\ =log\left|e^n-1\right|-log{e}^x+C.\end{array}$

Putting x⁴ = tÞ 4x³dx = dt

$\begin{array}{l}\therefore {\displaystyle \int \frac{1}{x(x^4-1)}dx=\frac{1}{4}{\displaystyle \int \frac{dt}{t(t-1)}=\frac{1}{4}{\displaystyle \int \frac{t-(t-1)}{t(t-1)}}dt}}\\ =\frac{1}{4}\left\{{\displaystyle \int \frac{dt}{t-1}-{\displaystyle \int \frac{dt}{t}}}\right\}\\ =\frac{1}{4}\left\{log\left|t-1\right|-logt\right\}+\; C\end{array}$

$\begin{array}{l}=\frac{1}{4}log\left|\frac{t-1}{t}\right|+\; C\\ =\frac{1}{4}log\left|\frac{x^4-1}{x^4}\right|+\; C\end{array}$

Putting x² = t such that 2xdx = dt

$\begin{array}{l}{\displaystyle \int \frac{2xdx}{(x^2+1)(x^2+3)}={\displaystyle \int \frac{dt}{(t+1)(t+3)}=\frac{1}{2}{\displaystyle \int \frac{2dt}{(t+1)(t+3)}}}}\\ =\frac{1}{2}{\displaystyle \int \frac{(t+3)-(t+1)dt}{(t+1)(t+3)}}\\ =\frac{1}{2}\left\{{\displaystyle \int \frac{dt}{t+1}}-\frac{dt}{t+3}\right\}\\ =\frac{1}{2}\left\{log\left|t+1\right|-log\left|t+3\right|\right\}+\; c\\ =\frac{1}{2}log\left|\frac{t+1}{t+3}\right|+\; c\end{array}$

Kindly go through the solution