Question

If $\int \frac{(\sqrt{x}{)}^5}{(\sqrt{x}{)}^7+x^6}dx=alog\left(\frac{x^k}{1+x^k}\right)+c$, then a and k are

• A. $\frac{2}{5},\frac{5}{2}$
• B. $\frac{1}{5},\frac{2}{5}$
• C. $\frac{5}{2},\frac{1}{2}$
• D. $\frac{2}{5},\frac{1}{2}$

Answer 1

The correct option is A $\frac{2}{5},\frac{5}{2}$

Given

$\int \frac{(\sqrt{x}{)}^5}{(\sqrt{x}{)}^7+x^6}dx=alog\left(\frac{x^k}{1+x^k}\right)+c$

L.H.S Taking $x^6$ common from denominator

$\Rightarrow \int \frac{(\sqrt{x}{)}^5}{x^6(\frac{(\sqrt{x}{)}^7}{x^6}+1)}dx$

$\Rightarrow \int \frac{1}{\left(\sqrt{x}{)}^7\right(\frac{1}{\sqrt{x}}{)}^5+1}dx$

Now Let $\frac{1}{(\sqrt{x}{)}^5}=t$

$\frac{-5}{2(\sqrt{x}{)}^7}.dx=dt$

$\Rightarrow \int \frac{-2}{3}.\frac{1}{(t+1)}dt$

$\Rightarrow \frac{-2}{5}ln(t+1)+c$

put $t=\frac{1}{(\sqrt{x}{)}^5}$

$\Rightarrow \frac{-2}{5}ln\left(\frac{1+(\sqrt{x}{)}^5}{(\sqrt{x}{)}^5}\right)+c\Rightarrow \frac{2}{5}ln\left(\frac{(\sqrt{x}{)}^5}{1+(\sqrt{x}{)}^5}\right)+c$

on comparing $a=t\frac{2}{5}k=5/2$