Chapter 5.7, Problem 34E

To determine

To calculate: The value of given integral by making a substitution to express the integrand as a rational function

Answer to Problem 34E

The value of the given integral is $\frac{1}{2}\left[\ln \left(\sqrt{x-3}-1\right)+3\ln \left(\sqrt{x-3}+3\right)\right]+C$

Explanation of Solution

Given information:

The integral is ${\displaystyle \int \frac{dx}{2\sqrt{x+3}+x}}$

Formula used:

  ${\displaystyle \int \frac{1}{x}dx=\ln \left|x\right|+C}$

Calculation:

The given integral is,

  ${\displaystyle \int \frac{dx}{2\sqrt{x+3}+x}}$

Assume, $u=\sqrt{x+3}$ so

  $\begin{array}{l}{u}^2=x+3\\ x=u^2-3\\ d\left(u^2\right)=d\left(x+3\right)\left[\text{differentiate both sides}\right]\\ 2udu=dx\end{array}$

Now the given integral becomes,

  ${\displaystyle \int \frac{2u}{u^2+2u-3}du}$

Next the integral term is written as,

  $\frac{2u}{u^2+2u-3}=\frac{2u}{\left(u-1\right)\left(u+3\right)}=\frac{A}{u-1}+\frac{B}{u+3}$

Here, $A\text{ and }B$ are the two constants. To find the value of $A\text{ and }B$ multiply both sides by $\left(u-1\right)\left(u+3\right)$

  $\begin{array}{l}2u=A\left(u+3\right)+B\left(u-1\right)\\ 2u=Au+3A+Bu-B\\ 2u=u\left(A+B\right)+3A-B\end{array}$

The coefficient of $x$ must be equal and the constant terms are all equal. So

  $A+B=2\text{ and 3}A-B=0$

Solving this system of linear equations of $A\text{ and }B$ the value is found as,

  $\begin{array}{l}A+B=2\text{ and 3}A-B=0\\ A+B=2\Rightarrow B=2-A\\ -B=-3A\Rightarrow -2+A=-3A\Rightarrow A=\frac{1}{2},B=\frac{1}{2}\end{array}$

So, $A=\frac{1}{2}\text{ and }B=\frac{1}{2}$

Now, the integral is formed the term,

  ${\displaystyle \int \left[\frac{1}{2\left(u-1\right)}+\frac{3}{2\left(u+3\right)}\right]}du$

Simplify the integral and substitute $u$ for $\sqrt{x+3}$

  $\begin{array}{l}{\displaystyle \int \left[\frac{1}{2\left(u-1\right)}+\frac{3}{2\left(u+3\right)}\right]}du\\ =\frac{1}{2}\ln \left\{2\left(u-1\right)\right\}+\frac{3}{2}\ln \left\{2\left(u+3\right)\right\}+C\\ =\frac{1}{2}\left[\ln \left(u-1\right)+3\ln \left(u+3\right)\right]+C\\ =\frac{1}{2}\left[\ln \left(\sqrt{x-3}-1\right)+3\ln \left(\sqrt{x-3}+3\right)\right]+C\end{array}$

Therefore, the value of the given integral is $\frac{1}{2}\left[\ln \left(\sqrt{x-3}-1\right)+3\ln \left(\sqrt{x-3}+3\right)\right]+C$