Answer:
[tex]\int{\frac{\sqrt{x + 49}}{x}} \, dx =2\sqrt{x + 49} +49\ln|\frac{\sqrt{x + 49}-7}{\sqrt{x + 49}+7}|+c[/tex]
Step-by-step explanation:
Given
[tex]\int\limits {\frac{\sqrt{x + 49}}{x}} \, dx[/tex]
Required
Solve by substitution
Let:
[tex]u = \sqrt{x + 49}[/tex]
Square both sides
[tex]u^2 = x + 49[/tex]
Differentiate
[tex]2udu = dx[/tex]
Also notice that:
[tex]x = u^2 - 49[/tex]
So, we have:
[tex]\int\limits {\frac{\sqrt{x + 49}}{x}} \, dx =\int\limits {\frac{u}{u^2 - 49}} \, 2udu[/tex]
[tex]\int\limits {\frac{\sqrt{x + 49}}{x}} \, dx =\int\limits {\frac{2u^2}{u^2 - 49}} \, du[/tex]
Add 0 to the numerator
[tex]\int\limits {\frac{\sqrt{x + 49}}{x}} \, dx =\int\limits {\frac{2u^2+0}{u^2 - 49}} \, du[/tex]
Express 0 as 98 -98
[tex]\int\limits {\frac{\sqrt{x + 49}}{x}} \, dx =\int\limits {\frac{2u^2-98+98}{u^2 - 49}} \, du[/tex]
Split the fraction
[tex]\int{\frac{\sqrt{x + 49}}{x}} \, dx =\int( \frac{2u^2-98}{u^2 - 49}+\frac{98}{u^2 - 49} )\, du[/tex]
[tex]\int{\frac{\sqrt{x + 49}}{x}} \, dx =\int( \frac{2(u^2-49)}{u^2 - 49}+\frac{98}{u^2 - 49} )\, du[/tex]
[tex]\int{\frac{\sqrt{x + 49}}{x}} \, dx =\int( 2+\frac{98}{u^2 - 49} )\, du[/tex]
Rewrite as:
[tex]\int{\frac{\sqrt{x + 49}}{x}} \, dx =\int( 2+\frac{98}{u^2 - 7^2} )\, du[/tex]
Integrate
[tex]\int{\frac{\sqrt{x + 49}}{x}} \, dx =2u +\int\frac{98}{u^2 - 7^2} \, du[/tex]
Remove constant (98)
[tex]\int{\frac{\sqrt{x + 49}}{x}} \, dx =2u +98\int\frac{1}{u^2 - 7^2} \, du[/tex]
As a general rule,
[tex]\int \frac{1}{x^2 - a^2} \, dx = \frac{1}{2}\ln|\frac{x-a}{x+a}|[/tex]
So, we have:
[tex]\int \frac{1}{u^2 - 7^2} \, du = \frac{1}{2}\ln|\frac{u-7}{u+7}|[/tex]
[tex]\int{\frac{\sqrt{x + 49}}{x}} \, dx =2u +98\frac{1}{u^2 - 7^2} \, du[/tex] becomes
[tex]\int{\frac{\sqrt{x + 49}}{x}} \, dx =2u +98*\frac{1}{2}\ln|\frac{u-7}{u+7}|+c[/tex]
[tex]\int{\frac{\sqrt{x + 49}}{x}} \, dx =2u +49\ln|\frac{u-7}{u+7}|+c[/tex]
Substitute values for u
[tex]\int{\frac{\sqrt{x + 49}}{x}} \, dx =2\sqrt{x + 49} +49\ln|\frac{\sqrt{x + 49}-7}{\sqrt{x + 49}+7}|+c[/tex]
