Chapter G, Problem 36E

To determine

To calculate : ${\displaystyle {\int }_{}^{}\frac{dx}{2\sqrt{x+3}+x}}$

Answer to Problem 36E

  $\ln \left(\left|2\sqrt{x+3}+x\right|\right)-\frac{\ln \left(\left|\sqrt{x+3}-1\right|\right)}{2}+\frac{\ln \left(\sqrt{x+3}+3\right)}{2}+C$

Explanation of Solution

Given information :

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

Formula used :

Partial fraction formula:

  $\frac{px+q}{\left(x-a\right)\left(x-b\right)}=\frac{A}{x-a}+\frac{B}{x-b},a\ne b$

  ${\displaystyle {\int }_{}^{}\frac{1}{x}}dx=\ln \left(x\right)$

Calculation:

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

Apply u-substitution:

  $u=\sqrt{x+3}$

Differentiate with respect to ‘x’:

  $\frac{du}{dx}=\frac{1}{2\sqrt{x+3}}\Rightarrow dx=2\sqrt{x+3}du$ , use:

  $x=u^2-3$

So, the integration becomes:

  $\begin{array}{l}{\displaystyle {\int }_{}^{}\frac{dx}{2\sqrt{x+3}+x}}={\displaystyle {\int }_{}^{}\frac{1}{2u+u^2-3}\times 2\sqrt{u^2-3+3}du}\\ =2{\displaystyle {\int }_{}^{}\frac{u}{u^2+2u-3}du}\end{array}$

Write $u$ as $\frac{1}{2}\left(2u+2\right)-1$ and split:

  $\begin{array}{l}\Rightarrow {\displaystyle {\int }_{}^{}\frac{u}{u^2+2u-3}du}={\displaystyle {\int }_{}^{}\left(\frac{2u+2}{2\left(u^2+2u-3\right)}-\frac{1}{u^2+2u-3}\right)du}\\ ={\displaystyle {\int }_{}^{}\frac{2u+2}{2\left(u^2+2u-3\right)}du}-{\displaystyle {\int }_{}^{}\frac{1}{u^2+2u-3}}du\\ ={\displaystyle {\int }_{}^{}\frac{u+1}{u^2+2u-3}du}-{\displaystyle {\int }_{}^{}\frac{1}{u^2+2u-3}}du\mathrm{.....}\left(1\right)\end{array}$

Now solving:

  ${\displaystyle {\int }_{}^{}\frac{u+1}{u^2+2u-3}du}$

Apply v-substitution:

  $v=u^2+2u-3$

Differentiate with respect to ‘u’:

  $\frac{dv}{dx}=2u+2\to du=\frac{1}{2\left(u+1\right)}dv$

So, the integration becomes:

  $\begin{array}{l}{\displaystyle {\int }_{}^{}\frac{u+1}{u^2+2u-3}du}={\displaystyle {\int }_{}^{}\frac{u+1}{v}\times \frac{1}{2\left(u+1\right)}dv}\\ =\frac{1}{2}{\displaystyle {\int }_{}^{}\frac{1}{v}}dv\\ =\frac{\ln \left(v\right)}{2}\end{array}$

Substitute back: $v=u^2+2u-3$

  ${\displaystyle {\int }_{}^{}\frac{u+1}{u^2+2u-3}du}=\frac{\ln \left(u^2+2u-3\right)}{2}\mathrm{....}\left(2\right)$

Now solving:

  ${\displaystyle {\int }_{}^{}\frac{1}{u^2+2u-3}du}$

Factor the denominator:

  ${\displaystyle {\int }_{}^{}\frac{1}{\left(u-1\right)\left(u+3\right)}du}$

Apply partial fraction decomposition:

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

Write the right-hand side as a single fraction:

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

The denominators are equal, so we require the equality of the numerators:

  $1=A\left(u+3\right)+\left(u-1\right)B$

Expand the right-hand side:

  $1=uA+uB+3A-B$

Collect up the like terms:

  $1=u\left(A+B\right)+3A-B$

The coefficients near the like terms should be equal, so the following system is obtained:

  $\begin{array}{l}A+B=0\Rightarrow A=-B\mathrm{....}\left(a\right)\\ 3A-B=1\mathrm{....}\left(b\right)\end{array}$

From (a) and (b), we get

  $\begin{array}{l}3\left(-B\right)-B=1\\ \Rightarrow -3B-B=1\\ \Rightarrow \boxed{B=-\frac{1}{4}}\\ \Rightarrow \boxed{A=\frac{1}{4}}\end{array}$

Therefore,

  $\begin{array}{l}{\displaystyle {\int }_{}^{}\frac{1}{u^2+2u-3}du}={\displaystyle {\int }_{}^{}\left(\frac{1}{4\left(u-1\right)}-\frac{1}{4\left(u+3\right)}\right)}du\\ =\frac{1}{4}{\displaystyle {\int }_{}^{}\frac{1}{u-1}du}-\frac{1}{4}{\displaystyle {\int }_{}^{}\frac{1}{u+3}}du\end{array}$

Apply substitution: $v=u-1\Rightarrow \frac{dv}{du}=1\Rightarrow du=dv$

  $w=u+3\Rightarrow \frac{dw}{du}=1\Rightarrow du=dw$

So, the integration becomes:

  $\begin{array}{l}{\displaystyle {\int }_{}^{}\frac{1}{u^2+2u-3}du}=\frac{1}{4}{\displaystyle {\int }_{}^{}\frac{1}{v}dw}-\frac{1}{4}{\displaystyle {\int }_{}^{}\frac{1}{w}}dw\\ =\frac{\ln v}{4}-\frac{\ln w}{4}\end{array}$

Substitute back: $v=u-1,w=u+3$

  ${\displaystyle {\int }_{}^{}\frac{1}{u^2+2u-3}du}=\frac{\ln \left(u-1\right)}{4}-\frac{\ln \left(u+3\right)}{4}\mathrm{......}\left(3\right)$

From (1), (2) and (3), we have

  $\begin{array}{l}{\displaystyle {\int }_{}^{}\frac{u}{u^2+2u-3}du}=\frac{\ln \left(u^2+2u-3\right)}{2}-\frac{\ln \left(u-1\right)}{4}+\frac{\ln \left(u+3\right)}{4}\\ \Rightarrow 2{\displaystyle {\int }_{}^{}\frac{u}{u^2+2u-3}du}=\ln \left(u^2+2u-3\right)-\frac{\ln \left(u-1\right)}{2}+\frac{\ln \left(u+3\right)}{2}\end{array}$

Substitute back: $u=\sqrt{x+3}$

  $\begin{array}{l}{\displaystyle {\int }_{}^{}\frac{dx}{2\sqrt{x+3}+x}}=\ln \left({\left(\sqrt{x+3}\right)}^2+2\left(\sqrt{x+3}\right)-3\right)-\frac{\ln \left(\sqrt{x+3}-1\right)}{2}+\frac{\ln \left(\sqrt{x+3}+3\right)}{2}+C\\ {\displaystyle {\int }_{}^{}\frac{dx}{2\sqrt{x+3}+x}}=\ln \left(\left|2\sqrt{x+3}+x\right|\right)-\frac{\ln \left(\left|\sqrt{x+3}-1\right|\right)}{2}+\frac{\ln \left(\sqrt{x+3}+3\right)}{2}+C\end{array}$