Integrals

$\begin{array}{l}=sin\left(ax+b\right)cos\left(ax+b\right)=\frac{2sin\left(ax+b\right)cos\left(ax+b\right)}{2}\\ =\frac{sin2(ax+b)}{2}\\ Put2\left(ax+b\right)=t\\ 2adx=dt\\ I=\int \frac{sin2(ax+b)}{2}dx\\ =\frac{1}{2}\int \frac{sintdt}{2a}=\frac{1}{4a}[-cost]+C\\ =-\frac{1}{4a}cos2(ax+b)+C\end{array}$

Kindly go through the solution

$\begin{array}{l}\frac{x(1+logx)}{}\\ Put,1+logx=t\\ =\frac{1dx}{x}=dt\\ =I=\int \frac{1}{x+xlogx}dx=\int \frac{1}{t}dt\\ =log\left(t\right)+c\\ =log\left(1+logx\right)+c\end{array}$

$\begin{array}{l}Let,logx=t\\ \frac{1}{x}dx=dt\\ I=\int \frac{{(logx)}^2}{x}dx=\int t^{2}dt=\frac{t^3}{3}+C\\ =\frac{{(logx)}^3}{3}+C\end{array}$

$\begin{array}{l}Let,1+x^2=t\\ \Rightarrow 2xdx=dt\\ I=\int \frac{2x}{1+x^2}dx=\int \frac{1dt}{t}\\ \Rightarrow log\left|t\right|+c\\ \Rightarrow log\left|1+x^2\right|+c\end{array}$

$\begin{array}{l}f\left(x\right)={\displaystyle \int 4x^3}–\frac{3}{x^4}dx\\ =4\cdot \frac{x^4}{4}-3\cdot \frac{x^{-3}}{-3}+\; C\\ =x^4+\frac{1}{x^3}+\; C\end{array}$

Now,  $f\left(2\right)=0 \therefore f\left(2\right) =2^4+\frac{1}{2^3}+C=0$

$\begin{array}{l}=16+\frac{1}{8}+c=0\\ =c=-\left(16+\frac{1}{8}\right)\\ =c=-\frac{129}{8}\end{array}$

Therefore, correct answer is A.

$\frac{x^{3/2}}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{x^{1/2}}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\; C$

$=\frac{2}{3}{x}^{3/2}+2x^{1/2}+$ C

? The correct Answer is (C)

$\begin{array}{l}\int \left(\frac{2}{co{s}^{2}x}-\frac{3sinx}{co{s}^{2}x}\right)dx\\ \begin{array}{l}={\displaystyle \int \left(2se{c}^{2}x–3tanx.secx\right)dx}\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}={\displaystyle \int 2se{c}^{2}x}.dx–{\displaystyle \int 3tan}x.secx.dx\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}=2tanx– 3 secx+C\hfill \end{array}\end{array}$

$\begin{array}{l}\int \frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$co{s}^2$}\right.x}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$co{s}^2$}\right.x}dx\\ =\int \frac{si{n}^{2}x}{co{s}^{2}x}dx\\ \begin{array}{l}={{\displaystyle \int tan}}^{2}xdx\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}={\displaystyle \int \left(se{c}^{2}x–1\right)}dx={\displaystyle \int se{c}^2}xdx–{\displaystyle \int 1}.dx\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}=tanx–x+C\hfill \end{array}\end{array}$

$={\displaystyle \int \left(se{c}^{2}x+secxtanx\right)dx}$

$={\displaystyle \int se{c}^{2}x.dx}+{\displaystyle \int secxtanxdx}$

$=tanx+secx+C$