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Evaluate the integral using substitution

by RoRi

Use the method of substitution to evaluate the following integral:

    \[ \int (x^2+1)^{-\frac{3}{2}} \, dx. \]

Let

    \begin{align*} u &= (x^2+1)^{-\frac{1}{2}} \quad \implies \quad du = -x(x^2+1)^{-\frac{3}{2}} \\ x &= \frac{\sqrt{1-u^2}}{u}. \end{align*}

Then, we can evaluate the integral,

    \begin{align*} \int (x^2+1)^{-\frac{3}{2}} \, dx &= \int -\frac{1}{x} \, du \\ &= \int \frac{-u}{\sqrt{1-u^2}} \, du. \end{align*}

Now, we make a second substitution. Let

    \[ v = 1-u^2 \quad \implies \quad dv = -2u \, du. \]

Then, we continue evaluating the integral,

    \begin{align*} \int \frac{-u}{\sqrt{1-u^2}} \, du &= \frac{1}{2} \int \frac{1}{\sqrt{v}} \, dv \\ &= \sqrt{v} + C \\ &= \sqrt{1-u^2} + C \\ &= \sqrt{1 - \frac{1}{x^2+1}} + C \\[9pt] &= \sqrt{\frac{x^2}{x^2+1}} + C \\[9pt] &= \frac{x}{\sqrt{x^2+1}} + C. \end{align*}

5 comments

  2. Johny Diala says:

    In my opinion, there are a few minor errors here. In the top of your solution, you have

        \[x=\frac{\sqrt{1 - u^2}}{u}.\]

    It should be

        \[x=\frac{±\sqrt{1 - u^2}}{u}\]

    Moreover, during the end, you assert

        \[\sqrt{\frac{x^2}{x^2 + 1}} = \frac{x}{\sqrt{x^2 + 1}}\]

    though this again presumes x\geq0.

    How do we deal with this? You can, firstly, only solve the integral for x\geq0 so your original u substitution is one-to-one. Now, since the integrand is even, it follows that the primitive is odd, and hence we have a solution

        \[\sqrt{\frac{x^2}{x^2 + 1}} + C\]

    for x \geq 0 and

        \[-\sqrt{\frac{x^2}{x^2 + 1}} + C\]

    for x<0. With this, one can simplify the integral to

        \[\frac{x}{\sqrt{x^2 + 1}} + C\]

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