Question

$\int \frac{(\sqrt{x}{)}^3}{(\sqrt{x}{)}^5+x^4}dx=A\log \left(\frac{x^k}{x^k+1}\right)+c$ then the values of $A$ and $k$ respectively are.

• A. $\frac{3}{2}\&\frac{2}{3}$
• B. $\frac{3}{2}$ & $2$
• C. does not exist
• D. $\frac{2}{3}\&\frac{3}{2}$

Answer 1

The correct option is D $\frac{2}{3}\&\frac{3}{2}$

$\int \frac{(\sqrt{x}{)}^3}{(\sqrt{x}{)}^5+x^4}dx=A\log \left(\frac{x^k}{x^k+1}\right)+c$

Let $I=\int \frac{(\sqrt{x}{)}^3}{(\sqrt{x}{)}^5+x^4}dx$

Put $\sqrt{x}=t\Rightarrow dx=2tdt$

$I=2\int \frac{dt}{t+t^4}$

$=2\int \frac{dt}{t(1+t)(t^2-t+1)}$

We will use partial dfraction

$\frac{1}{t(1+t)(t^2-t+1)}=\frac{A}{t}+\frac{B}{t+1}+\frac{Cx+D}{t^2-t+1}$

$1=A(1+t)(t^2-t+1)+Bt(t^2-t+1)+(Ct+D)(t^2+t)$

For $t=-1\Rightarrow B=-\frac{1}{3}$

For $t=0\Rightarrow A=1$

For $t=1\Rightarrow C+D=-\frac{1}{3}$

For $t=2\Rightarrow 2C+D=-1$

$\Rightarrow D=\frac{1}{3},C=-\frac{2}{3}$

$I=2log\left|t\right|-\frac{2}{3}log|t+1|-\frac{2}{3}log|t^2-t+1|+c$

$I=2log\left|t\right|-\frac{2}{3}log|t^3+1|+c$

$I=\frac{2}{3}log\left|t^3\right|-\frac{2}{3}log|t^3+1|+c$

$I=\frac{2}{3}log\left|\frac{t^3}{t^3+1}\right|+C$

$I=\frac{2}{3}log\left|\frac{x^{3/2}}{x^{3/2}+1}\right|+C$

On comparing, $\Rightarrow A=\frac{2}{3},k=\frac{3}{2}$