Indefinite integral : $\int \frac{dx}{(1+x^{2})^{\frac{3}{2}}}$

Question

I am stuck in this exercise of calculus about solving this indefinite integral, so I would like some help from your part:

$$\int \frac{dx}{(1+x^{2})^{\frac{3}{2}}}$$

Answer 1

Hint 1: $\cosh^2(x)-\sinh^2(x)=1$ and use change of variables.

Hint 2:

$$\frac{\mathrm{d}}{\mathrm{d}x}\left(\tanh(x)\right) =\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\sinh(x)}{\cosh(x)}\right) = \frac{1}{\cosh^2(x)}\left(\cosh^2(x)-\sinh^2(x)\right) = \text{sech}^2(x)$$

Answer 2

We have $t>0$.

Recall that

$$\int\frac1{(t^2-x^2)^{1/2}}\ dx=\arcsin\left(\frac xt\right)+c_1$$

Take the derivative of both sides with respect to $t$ to get

$$\int\frac{-t}{(t^2-x^2)^{3/2}}\ dx=\frac{-x}{|t|\sqrt{t^2-x^2}}+c_2$$

Setting $t=1$ and simplifying, we reach

$$\int\frac1{(1-x^2)^{3/2}}\ dx=\frac x{\sqrt{1-x^2}}+c_3$$

Answer 3

Let $x = \sinh u$. Then the integral becomes $$\int \frac{\mathrm dx}{(1+x^{2})^{\frac{3}{2}}} = \int \frac{\cosh u}{\cosh^3 u}\mathrm du = \int\operatorname{sech}^2 u\,\mathrm du$$

Now, $\operatorname{sech}^2 u$ is the derivative of a well-known hyperbolic function. Can you spot it?

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Previous answer, when the question had a minus in the denominator

Let $x = \sin u$. Then we have $$\int \frac{\mathrm dx}{(1-x^{2})^{\frac{3}{2}}} = \int \frac{\cos u}{\cos^3 u}\mathrm du = \int\sec^2 u\,\mathrm du$$

Now, $\sec^2 u$ is the derivative of a well-known trigonometric function. Can you spot it?

Then, if you desire to remove trigonometric functions, recall the identity $\sin^2 x + \cos^2 x = 1$ and you're done.