Find the integral \int \frac{x-3}{\sqrt{x^2 + 1}} \,dx

Question:

Find the integral

{eq}\int \frac{x-3}{\sqrt{x^2 + 1}} \,dx {/eq}

Indefinite Integral:

The partial fractional method is used for the long integrand expression. Then two or more parts of the integral is solved individually, by applying the various rules. The result of integral is added with the integration constant at the end of the expression.

Answer and Explanation:

The indefinite integral to be solved here is:

{eq}\int \frac{x-3}{\sqrt{x^2 + 1}} \,dx {/eq}

So we will first break it in two parts as follows:

{eq}\int \frac{x-3}{\sqrt{x^2 + 1}} \,dx=\int \frac{x}{\sqrt{x^2+1}}dx-\int \frac{3}{\sqrt{x^2+1}}dx\\ {/eq}

Next, we will solve the integral:

{eq}\int \frac{x}{\sqrt{x^2+1}}dx=\sqrt{x^2+1}\\ {/eq}

using the substitution of:

{eq}u=x^2+1\\ \Rightarrow \frac{du}{dx}=2x\\ {/eq}

So the indefinite integral is now;

{eq}\int \frac{x}{\sqrt{x^2+1}}dx=\sqrt{x^2+1}=\int \frac{1}{2\sqrt{u}}du\\ =\frac{1}{2}\cdot \frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c~~~~~~~~~~~~\left [ \because \int x^adx=\frac{x^{a+1}}{a+1}+c \right ]\\ =\sqrt{x^2+1}+c~~~~~~~~~~~~~~~\left [ \because u=x^2+1 \right ] {/eq}

Now the integral:

{eq}\int \frac{3}{\sqrt{x^2+1}}dx=3\ln \left|\sqrt{x^2+1}+x\right|+c~~~~~~~~~~\left [ \because \int \frac{1}{\sqrt{x^2+1}}dx=\ln \left|\sqrt{x^2+1}+x\right| +c \right ] \\ {/eq}

So finally adding the two results we will have:

{eq}\int \frac{x}{\sqrt{x^2+1}}dx-\int \frac{3}{\sqrt{x^2+1}}dx\\ =\sqrt{x^2+1}-3\ln \left|\sqrt{x^2+1}+x\right|+C\\ \Rightarrow \int \frac{x-3}{\sqrt{x^2+1}}dx=\sqrt{x^2+1}-3\ln \left|\sqrt{x^2+1}+x\right|+C {/eq}