Q. 26 - Calculus

Question

Indefinite integrals: Use integration formulas, algebra, and educated guess-and-check strategies to find the following integrals. 

$\int \frac{1}{1-2x^2}dx$

Step-by-Step Solution

Answer

The solution is: $-\frac{1}{2\sqrt{2}}\left[\ln \left|\sqrt{2}x+1\right|-\ln \left|\sqrt{2}x-1\right|\right]+C$

Step 1: Given

The given integral is:  $\int \frac{1}{1-2x^2}dx$

Step 2: To Find

Use integration formulas, algebra, and educated guess-and-check strategies to find the given integral. 

Step 3: Calculation

Let,

$$\begin{gathered} u=\sqrt{2}x \\ du=\sqrt{2}dx \\ \frac{du}{\sqrt{2}}=dx \end{gathered}$$

Using substitution we get:

$$\begin{gathered} \int \frac{1}{1-2x^2}dx \\ =\int \frac{1}{1-{\left(\sqrt{2}x\right)}^2}dx \\ =\int \frac{1}{1-u^2}\frac{du}{\sqrt{2}} \\ =\frac{1}{\sqrt{2}}\int \frac{1}{1-u^2}du \\ =-\frac{1}{\sqrt{2}}\int \frac{1}{u^2-1}du \\ =-\frac{1}{\sqrt{2}}\int \frac{1}{\left(u+1\right)\left(u-1\right)}du \\ =-\frac{1}{2\sqrt{2}}\int \frac{2}{\left(u+1\right)\left(u-1\right)}du \\ =-\frac{1}{2\sqrt{2}}\int \frac{\left(u+1\right)-\left(u-1\right)}{\left(u+1\right)\left(u-1\right)}du \\ =-\frac{1}{2\sqrt{2}}\int \left[\frac{1}{u-1}-\frac{1}{u+1}\right]du \\ =-\frac{1}{2\sqrt{2}}\left[\ln \left|u+1\right|-\ln \left|u-1\right|\right]+C \\ =-\frac{1}{2\sqrt{2}}\left[\ln \left|\sqrt{2}x+1\right|-\ln \left|\sqrt{2}x-1\right|\right]+C \end{gathered}$$

Where, C = integrating constant.