Question

The value of the integral $\int \frac{x^{2} - 4 x \sqrt{x} + 6 x - 4 \sqrt{x} + 1}{x - 2 \sqrt{x} + 1} d x$ 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)

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