Question

Evaluate the following:

${\displaystyle \int \frac{x}{\sqrt{x}+1}\text{d}x}$  (Hint: Put  ${\displaystyle \sqrt{x}}$ = z)

Answer 1

I = ${\displaystyle \int \frac{x}{\sqrt{x}+1}\text{d}x}$ 

Put  ${\displaystyle \sqrt{x}}$ = t

⇒ x = t²

∴ dx = 2t . dt

∴ I = ${\displaystyle \int \frac{\text{t}\cdot 2\text{t}\cdot \text{dt}}{\text{t}+1}}$

= ${\displaystyle 2\int \frac{{\text{t}}^3}{\text{t}+1}\text{dt}}$

= ${\displaystyle 2\int \frac{{\text{t}}^3+1-1}{\text{t}+1}\text{dt}}$

= ${\displaystyle 2\int \frac{{\text{t}}^3+1}{\text{t}+1}\text{dt}-2\int \frac{1}{\text{t}+1}\text{dt}}$

= ${\displaystyle 2\int \frac{(\text{t}+1)({\text{t}}^2-\text{t}+1)}{\text{t}+1}\text{dt}-2\int \frac{1}{\text{t}+1}\text{dt}}$

= ${\displaystyle 2\int ({\text{t}}^2-\text{t}+1)\text{dt}-2\int \frac{1}{\text{t}+1}\text{dt}}$

= ${\displaystyle 2[\frac{{\text{t}}^3}{3}-\frac{{\text{t}}^2}{2}+\text{t}]-2\log |\text{t}+1|}$

= ${\displaystyle 2[\frac{x^{\frac{3}{2}}}{3}-\frac{x}{2}+\sqrt{x}]-2\log |\sqrt{x}+1|+\text{C}}$

= ${\displaystyle 2[\frac{x\sqrt{x}}{3}-\frac{x}{2}+\sqrt{x}-\log |\sqrt{x}+1|]+\text{C}}$

Hence, I = ${\displaystyle 2[\frac{x\sqrt{x}}{3}-\frac{x}{2}+\sqrt{x}-\log |\sqrt{x}+1|]+\text{C}}$