Question

The value of the integral $\int \frac{x^2-4x\sqrt{x}+6x-4\sqrt{x}+1}{x-2\sqrt{x}+1}dx$ is equal to ( where $c$ is constant of integration)

• A. $\frac{x^{\frac{3}{2}}}{2}+x+c$
• B. $\frac{x^2}{2}-\frac{4}{3}{x}^{\frac{3}{2}}+x+c$
• C. $x^{\frac{3}{2}}+\frac{x}{2}+c$
• D. $\frac{2}{3}{x}^{\frac{3}{2}}+c$

Answer 1

Answer: (B)

$I=\int \frac{{\left(\sqrt{x}-1\right)}^4}{{\left(\sqrt{x}-1\right)}^2}dx=\int {\left(\sqrt{x}-1\right)}^{2}dx=\int \left(x-2\sqrt{x}+1\right)dx$
$=\frac{x^2}{2}-\frac{4}{3}{x}^{\frac{3}{2}}+x+c$