Question

If $I=\int \frac{(2x-3{)}^{1/2}}{(2x-3{)}^{1/3}+1}dx$
$=3\left[\frac{1}{7}\right(2x-3{)}^{7/6}-\frac{1}{5}(2x-3{)}^{5/6}+\frac{1}{3}(2x-3{)}^{1/2}-(2x-3{)}^{1/6}+g(x\left)\right]+C$
then g(x) is equal to

• A. $ta{n}^{-1}(2x-3{)}^{1/6}$
• B. $(2x-3{)}^{1/2}$
• C. $3ta{n}^{-1}(2x-3{)}^{1/6}$
• D. $4(2x-3{)}^{1/6}$

Answer 1

The correct option is A $ta{n}^{-1}(2x-3{)}^{1/6}$

Let $I=\int \frac{\sqrt{2x-3}}{\sqrt[3]{32x-3}+1}dx$
Put $u=2x-3\Rightarrow du=2dx$
$I=\frac{1}{2}\int \frac{\sqrt{u}}{\sqrt[3]{3u}+1}du$
Put $s=\sqrt[6]{6u}\Rightarrow ds=\frac{1}{6u^{\frac{5}{6}}}du$
$\therefore I=3\int \frac{s^8}{s^2+1}ds=3\int (s^6-s^4+s^2+\frac{1}{s^2+1}-1)ds$
$=3\frac{s^7}{7}-3\frac{s^5}{5}+s^3-3s+3{\tan }^{-1}s+c$
$=3\frac{u^{\frac{7}{6}}}{7}-3\frac{u^{\frac{5}{6}}}{5}+\sqrt{u}-3\sqrt[6]{6u}+3{\tan }^{-1}\sqrt[6]{6u}+c$
$=\frac{3}{7}{(2x-3)}^{\frac{7}{6}}-\frac{3}{5}{(2x-3)}^{\frac{5}{6}}+\sqrt{2x-3}-3\sqrt[6]{62x-3}+3{\tan }^{-1}\sqrt[6]{62x-3}+c$