How do you integrate $\int x \sqrt{x^{2} + 9}$ $int xsqrt(x^2+9)$ using substitution?

Question

3357 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-01T06:39:55

The answer is $= \frac{1}{3} {(x^{2} + 9)}^{\frac{3}{2}} + C$ $=1/3(x^2+9)^(3/2)+C$

We need

$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C (n \neq - 1)$ $intx^ndx=x^(n+1)/(n+1)+C(n!=-1)$

Let $u = x^{2} + 9$ $u=x^2+9$

$d u = 2 x d x$ $du=2xdx$

$x d x = \frac{1}{2} d u$ $xdx=1/2du$

$\int x \sqrt{x^{2} + 9} d x = \frac{1}{2} \int \sqrt{u} d u$ $intxsqrt(x^2+9)dx=1/2intsqrtudu$

$= \frac{1}{2} \cdot \frac{u^{\frac{3}{2}}}{\frac{3}{2}}$ $=1/2*u^(3/2)/(3/2)$

$= \frac{1}{3} {(x^{2} + 9)}^{\frac{3}{2}} + C$ $=1/3(x^2+9)^(3/2)+C$

How do you integrate ∫x√x2+9int xsqrt(x^2+9) using substitution?