Question

Make a substitution to express the integrand as a rational function and then evaluate the integral. (Remember to use absolute values where appropriate. Use C for the constant of integration.)
integral sqrt(x+36)*dx / x

Answer 1

Answer:

$\mathbf{I =2{\sqrt{x+36}} + 6 \ In \begin {vmatrix} \dfrac{{\sqrt{x+36}}-6}{{\sqrt{x+36}}+6}\end {vmatrix}+ C}$

Step-by-step explanation:

Given that :

$I = \int \dfrac{\sqrt{x+36}}{x} \dx = \int \dfrac{2u^2}{u^2-36} \ du$

x = u² - 36

Let u = $\sqrt{x+36}$

Then:

$du =\dfrac{1}{2 \sqrt{x+36}}dx$

$2 \sqrt{x+36} \ \ du =dx$

$2 u^2 \ \ du =dx$

$\dfrac{2u^2}{u^2-36} \\ \\ 2u^2 \\ \\ 2u^2 - 72$

∴

$I = \int 2 du + \int \dfrac{72}{u^2-36}\ du$

$I = \int 2 du + 72 \int \dfrac{du}{u^2-6^2}$

$I =2u+ 72 \times\dfrac{1}{2\times 6} \ In \begin {vmatrix} \dfrac{u-6}{u+6}\end {vmatrix}+ C$

$I =2u + 6 \ In \begin {vmatrix} \dfrac{u-6}{u+6}\end {vmatrix}+ C$

substituting the value of u, we have:

$\mathbf{I =2{\sqrt{x+36}} + 6 \ In \begin {vmatrix} \dfrac{{\sqrt{x+36}}-6}{{\sqrt{x+36}}+6}\end {vmatrix}+ C}$