Question

If $\int \frac{(\sqrt{x}{)}^5}{(\sqrt{x}{)}^7+x^6}dx=a\log \left(\frac{x^k}{1+x^k}\right)+c$, 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 $\frac{2}{5},\frac{5}{2}$

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

$=\int \frac{x^{5/2}}{x^{7/2}+x^6}dx$

$=\int \frac{1}{x^{7/2-5/2}+x^{6-5/2}}dx$

$=\int \frac{1}{x^1+x^{7/2}}dx$

$=\int \frac{1}{x^{1/2}(1+x^{7/2-1/2})}dx$

$=\int \frac{1}{x(1+x^{7/1-1})}dx$

$=\int \frac{1}{x(1+x^{5/2})}dx$

Let $1+x^{5/2}=t$

$\Rightarrow x={(t-1)}^{2/5}$

On differentiating, we have

$\Rightarrow \frac{5}{2}{x}^{5/2-1}dx=dt=\frac{5}{2}{x}^{3/2}dx=dt$

$\Rightarrow \frac{5}{2}{(t-1)}^{3/5}dx=dt$

$=\int \frac{dt}{\frac{5}{2}{(t-1)}^{3/5}{(t-1)}^{2/5}}dx$

$=\frac{2}{5}\int \frac{dt}{{(t-1)}^{5/5}t}$

$=\frac{2}{5}\frac{dt}{(t-1)t}$

$=\frac{2}{5}\int \frac{1}{(t-1)}dt-\int \frac{1}{t}dt$

$=\frac{2}{5}\left(\log (t-1)-\log t\right)+c$

$=\frac{2}{5}\log \frac{(t-1)}{t}+c$

$=\frac{2}{5}\log \frac{x^{5/2}}{1+x^{5/2}}+c$

Hence, $a=\frac{2}{5},$ $k=\frac{5}{2}.$

Hence, the answer is $\frac{2}{5},\frac{5}{2}$