⇒Indefinite Integration

Correct Option is (A)

Let $I=\int \frac{x^3}{\sqrt{1+x^2}}dx=\int \frac{x^2\cdot x}{\sqrt{1+x^2}}dx$

Put $\sqrt{1+x^2}=t\Rightarrow 1+x^2=t^2\Rightarrow 2xdx=2tdt$

$\begin{array}{rl}& \text{ Put }\sqrt{1+{\mathrm{x}}^2}=\mathrm{t}\Rightarrow 1+{\mathrm{x}}^2={\mathrm{t}}^2\Rightarrow 2\mathrm{xdx}=2\mathrm{td}\\ & \therefore \mathrm{I}=\int \frac{({\mathrm{t}}^2-1)\mathrm{tdt}}{\mathrm{t}}=\int ({\mathrm{t}}^2-1)d\mathrm{t}=\frac{{\mathrm{t}}^3}{3}-\mathrm{t}+\mathrm{c}=\frac{{(1+{\mathrm{x}}^2)}^{\frac{3}{2}}}{3}-\sqrt{1+{\mathrm{x}}^2}+\mathrm{c}\end{array}$

Comparing with given data, $a=\frac{1}{3},b=-1\Rightarrow a+b=\frac{-2}{3}$

Correct Option is (B)

Let $I=\int {\sec }^{4}x{\tan }^{4}xdx$

Put $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$

$\therefore$ $I=\int {\sec }^{2}x({\sec }^{2}x)({\tan }^{4}x)dx$

$=\int (1+t^2)(t{)}^{4}dt=\int (t^4+t^6)dt=\frac{t^5}{5}+\frac{t^7}{7}+c=\frac{{\tan }^{5}x}{5}+\frac{{\tan }^{7}x}{7}+c$

Correct Option is (D)

We have, $\frac{1}{\sqrt{2}}{\tan }^{-1}\left[\frac{f(x)}{\sqrt{2}}\right]+c=\int \frac{1+x^2}{1+x^4}dx$ ...... (1)

Let $I=\int \frac{1+x^2}{1+x^4}dx$

Dividing numerator and denominator by ${\mathrm{x}}^2$, we get

$\begin{array}{rl}& I=\int \frac{(1+\frac{1}{x^2})}{(x^2+\frac{1}{x^2})}dx=\int \frac{(1+\frac{1}{x^2})}{{(x-\frac{1}{x})}^2+2}dx\\ & \text{ Put }x-\frac{1}{x}=t\Rightarrow (1+\frac{1}{x^2})dx=dt\\ & \therefore I=\int \frac{dt}{t^2+2}=\int \frac{dt}{(t{)}^2+(\sqrt{2}{)}^2}\\ & =\frac{1}{\sqrt{2}}{\tan }^{-1}\left[\frac{(\mathrm{t})}{\sqrt{2}}\right]+\mathrm{c}=\frac{1}{\sqrt{2}}{\tan }^{-1}\left[\frac{(x-\frac{1}{x})}{\sqrt{2}}\right]+c......(2)\\ & \end{array}$

From (1) and (2), $f(x)=x-\frac{1}{x}$