# How am I computing $\int \frac{x^{3}}{\sqrt{x^{2} + 9}}dx$ incorrectly?

$$\displaystyle\int \frac{x^{3}}{\sqrt{x^{2} + 9}}dx$$

doing a trig sub:

$$x = 3\tan(\theta)$$

$$dx = 3\sec^{2}(\theta)d\theta$$

$$\displaystyle\int \frac{(3\tan(\theta))^{3}}{\sqrt{3\tan(\theta)^{2} + 9}}d\theta$$

$$\displaystyle\int \frac{ ( 27\tan^{3}(\theta) } { \sqrt{ 3\tan(\theta)^{2} + 9 } }d\theta$$

$$27\displaystyle\int \frac{\tan^{3}(\theta)\cdot\sec^{2}(\theta)}{\sqrt{\sec^{2}(\theta)}}d\theta$$

$$27\displaystyle\int \tan(\theta)\cdot\tan(\theta)\cdot\sec(\theta)\cdot d(\theta)$$

• Your approach looks fine. Keep in mind the pythagorean identity for tan and sec, and this is a solvable integral. Not easy by any means, but solvable. Mar 31, 2017 at 0:07
• There don't need to be bounds @TheGreatDuck...it's indefinite. Careful to replace the $dx$ with $3\sec^2(\theta)d \theta$ in the first integral after the substitution. Mar 31, 2017 at 0:07
• Here are a few errors: $x^2=9\tan^2(\theta)$ (but you have another error later, so it doesn't matter), and somehow, you lose a $\tan(\theta)$ in the last step. Now, in the last step, you can replace $\tan^2(\theta)$ again and hopefully recognize what is in front of you. Or you might consider the substitution $u=x^2+9$ instead for your integral. Mar 31, 2017 at 0:13
• Also, you might consider using \tan for $\tan$ and \sec for $\sec$ and \cdot for $\cdot$ (multiplication). Mar 31, 2017 at 0:18
• Thanks! Didn't see that I lost that tan so then now I can do a trig integral: tan^{2}(x)*tan(x)*sec(x)*dx and that to (sec^{2} - 1)*tan(x)*sec(x)dx and then u-sub that (u^2 - 1) take the integral 27 [ sec^{3}(x) / 3 - sec(x) ] creating a triangle I get: 1/9( √(9+x^2) / 3)^3 ) - 27( √(9+x^2) / 3 ) + C .... Is this right or wrong?
– yre
Mar 31, 2017 at 0:29

$$\int u dv = uv - \int v du$$
$$u = x^{2}, \qquad dv = \frac{x }{\sqrt{x^{2}+9}}dx$$ $$du = 2x dx, \qquad v = \sqrt{x^{2}+9}$$
$$\int \frac{x^{3}}{\sqrt{x^{2}+9}}dx = \frac{1}{3} \left(x^2-18\right) \sqrt{x^2+9}$$