Question

$\int \frac{x^{3/2}+x^{5/2}+x}{x^2+2x^{1/2}+x+1}dx$ equals

• A. $\frac{2}{3}{x}^{3/2}+x-{\tan }^{-1}{x}^{1/4}+C$
• B. $\frac{2}{3}{x}^{5/2}-x+{\tan }^{-1}{x}^{1/2}+C$
• C. $\frac{2}{3}{x}^{1/2}-x-{\tan }^{-1}{x}^{1/4}+C$
• D. $\frac{2}{3}{x}^{3/2}-x+{\tan }^{-1}{x}^{1/4}+C$

Answer 1

The correct option is A $\frac{2}{3}{x}^{3/2}+x-{\tan }^{-1}{x}^{1/4}+C$

Let $I=\frac{x^{3/2}+x^{5/2}+x}{x^2+2x^{1/2}+x+1}=\frac{x(x^{1/2}+x^{3/2}+1)}{x^2+x^{1/2}+x+x^{1/2}+1+x^{3/2}}$

$=\frac{x(x^{1/2}+x^{3/2}+1)}{x^{1/2}(x^{3/2}+x^{1/2}+1)+1(x^{3/2}+x^{1/2}+1)}$

$=\frac{x(x^{3/2}+x^{1/2}+1)}{(x^{1/2}+1)(x^{3/2}+x^{1/2}+1)}=\frac{x-1+1}{\sqrt{x}+1}$

$=\frac{{\left(\sqrt{x}\right)}^2-{\left(1\right)}^2}{\sqrt{x}+1}+\frac{1}{\sqrt{x}+1}=(\sqrt{x}-1)+\frac{1}{(\sqrt{x}+1)}$

$\therefore \int Idx=\int ((\sqrt{x}-1)+\frac{1}{(\sqrt{x}+1)})dx$

$=\int (\sqrt{x}-1)dx+\int \frac{1}{\sqrt{x}+1}dx$

$=\frac{x^{3/2}}{3/2}-x+\int \frac{1}{{\left(x^{1/4}\right)}^2+{\left(1\right)}^2}dx=\frac{2}{3}{x}^{3/2}+x-{\tan }^{-1}{x}^{1/4}+C$