Question

Solve:
$\int x^5\sqrt{a^3+x^3}dx$

• A. $\frac{1}{3}[\frac{2}{3}{a}^3\sqrt{(a^3+x^3{)}^3}+\frac{4}{15}{a}^3\sqrt{(a^3+x^3{)}^3}]+C$.
• B. $\frac{1}{3}[\frac{2}{3}{a}^3\sqrt{(a^3+x^3{)}^3}+\frac{2}{5}{a}^3\sqrt{(a^3+x^3{)}^3}]+C$.
  
• C. $\frac{1}{3}[\frac{2}{3}{a}^3\sqrt{(a^3+x^3{)}^3}-\frac{4}{5}{a}^3\sqrt{(a^3+x^3{)}^3}]+C$.
  
• D. $\frac{1}{3}[-\frac{2}{3}{a}^3\sqrt{(a^3+x^3{)}^3}+\frac{2}{5}{a}^3\sqrt{(a^3+x^3{)}^3}]+C$.
  

Answer 1

The correct option is D $\frac{1}{3}[-\frac{2}{3}{a}^3\sqrt{(a^3+x^3{)}^3}+\frac{2}{5}{a}^3\sqrt{(a^3+x^3{)}^3}]+C$.

$\int x^5\sqrt{x^3+a^3}dx$

$u=x^3+a^3\to dx=\frac{1}{3x^2}du$

$=\frac{1}{3}\int (u^{3/2}-a^3\sqrt{u})du$

$=\frac{1}{3}\left[\int u^{3/2}du-a^3\int \sqrt{u}du\right]$

$=\frac{1}{3}[\frac{u^{5/2}}{5/2}-a^3\times \frac{u^{3/2}}{3/2}]$

$=\frac{1}{3}[\frac{2(x^3+a^3{)}^{5/2}}{5}-a^3\times \frac{2(x^3+a^3{)}^{3/2}}{3}]$

$=\frac{1}{3}[\frac{2\sqrt{(x^3+a^3{)}^5}}{5}-a^3\times \frac{2\sqrt{(x^3+a^3{)}^3}}{3}]$

$\int x^5\sqrt{x^3+a^3}dx$

$=\frac{1}{3}[-\frac{2}{3}{a}^3\sqrt{(a^3+x^3{)}^3}+\frac{2}{5}{a}^3\sqrt{(a^3+x^3{)}^3}]+C$.