Indefinite integral $\int x\sqrt{1+x}\mathrm{d}x$ using integration by parts

Question

$$\int x\sqrt{1+x}\mathrm{d}x$$

$v'=\sqrt{1+x}$

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

$u=x$

$u'=1$

$$\frac{2x}{3}(1+x)^{\frac{3}{2}}-\int\frac{2}{3}(1+x)^{\frac{3}{2}}=\frac{2x}{3}(1+x)^{\frac{3}{2}}-\frac{2}{3}*\frac{2}{5}(1+x)^{\frac{5}{2}}+c$$

result: $$\frac{2x}{3}(1+x)^{\frac{3}{2}}-\frac{4}{15}(1+x)^{\frac{5}{2}}+c$$

Deriving: $$x(1+x)^\frac{1}{2}-\frac{20}{30}(1+x)^{\frac{3}{2}}$$

Where did I get wrong?

Answer 1

You're correct.

$u = x$, $dv=\sqrt{x+1}dx$ $\Rightarrow$ $v=\frac{2}{3}(x+1)^{3/2}$, $du=dx$.

&

$$\int udv=uv-\int vdu$$

$$\Downarrow$$

SOLUTION:

$$\int x \sqrt{x+1}dx=x\frac{2}{3}(x+1)^{3/2}-\int\frac{2}{3}(x+1)^{3/2}dx=\bbox[5px,border:2px solid #F0A]{x\frac{2}{3}(x+1)^{3/2} -\frac{4}{15}(x+1)^{5/2}+C}$$

VERIFICATION:

Do not forget applying the product rule when taking derivative.

$$\frac{d}{dx}\bigg(x\frac{2}{3}(x+1)^{3/2} -\frac{4}{15}(x+1)^{5/2}+C\bigg)=\frac{2}{3}(x+1)^{3/2}+x\sqrt{x+1}-\frac{2}{3}(x+1)^{3/2}+0=\bbox[5px,border:2px solid #F0A]{x\sqrt{x+1}}$$