How would I integrate $(x^2+1)^{\frac{3}{2}}$?

Question

I need to find $\int(x^2+1)^{\frac{3}{2}}dx.$ I started by trying to split it into $\int (x^2+1)(x^2+1)^{\frac{1}{2}}dx$ and then integrating by parts but that didn't seem to be working out. Is there a better way to go about it?

Answer 1

Try the substitution $x = \tan(\theta)$. Then $x^2 + 1 = \tan^2(\theta) + 1 = \sec^2(\theta)$.

Since $dx = \sec^2(\theta) \ d\theta$, then after simplifying your integral becomes:

$$\int \sec^5(\theta) d \theta$$

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Alternatively, you can try a hyperbolic trig substitution that, e.g., user17762 suggests.

Answer 2

Setting $x=\sinh(t)$, we obtain $$I = \int (x^2+1)^{3/2}dx = \int \cosh^3(t) \cosh(t)dt = \int \cosh^4(t)dt = \int \left(\dfrac{e^t + e^{-t}}2\right)^4dt$$ This gives us $$16I = \int \left(e^{4t} + 4e^{2t} + 6 + 4e^{-2t} + e^{-4t}\right)dt = 2 \int \cosh(4t)dt + 8 \int \cosh(2t)dt + 6t + \text{const}$$ I trust you can finish from here making use of the fact that $$\int \cosh(at)dt = \dfrac{\sinh(at)}a$$

Answer 3

To carry out your original idea, proceed as follows: $$ \int (1+x^2)^{n/2} \, dx = \int x^2 (1+x^2)^{n/2-1} \, dx + \int (1+x^2)^{(n-2)/2} \, dx $$ Integrating the first by parts, $$ \int x^2 (1+x^2)^{n/2-1} \, dx = \frac{1}{n}x(1+x^2)^{n/2} -\frac{1}{n} \int (1+x^2)^{n/2} \, dx, $$ so rearranging gives $$ \int (1+x^2)^{n/2} \, dx = \frac{1}{n+1}x(1+x^2)^{n/2} + \frac{n}{n+1}\int (1+x^2)^{(n-2)/2} \, dx $$ Applying this a couple of times to your integral gets us to $$ \int (1+x^2)^{3/2} \, dx = \frac{1}{4}x(1+x^2)^{3/2} + \frac{3}{4} \left( \frac{1}{2}x(1+x^2)^{1/2} + \frac{1}{2} \int \frac{dx}{\sqrt{1+x^2}} \right), $$ and the last integral is equal to $ \arg\sinh{x} $, basically by the inverse function rule.