# Make a substitution to express the integrand as a rational function and then evaluate the integral. (Remember to use absolute values where a

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.) x 49 x dx

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1 year 2021-09-03T04:10:45+00:00 1 Answers 14 views 0

$$\int{\frac{\sqrt{x + 49}}{x}} \, dx =2\sqrt{x + 49} +49\ln|\frac{\sqrt{x + 49}-7}{\sqrt{x + 49}+7}|+c$$

Step-by-step explanation:

Given

$$\int\limits {\frac{\sqrt{x + 49}}{x}} \, dx$$

Required

Solve by substitution

Let:

$$u = \sqrt{x + 49}$$

Square both sides

$$u^2 = x + 49$$

Differentiate

$$2udu = dx$$

Also notice that:

$$x = u^2 – 49$$

So, we have:

$$\int\limits {\frac{\sqrt{x + 49}}{x}} \, dx =\int\limits {\frac{u}{u^2 – 49}} \, 2udu$$

$$\int\limits {\frac{\sqrt{x + 49}}{x}} \, dx =\int\limits {\frac{2u^2}{u^2 – 49}} \, du$$

$$\int\limits {\frac{\sqrt{x + 49}}{x}} \, dx =\int\limits {\frac{2u^2+0}{u^2 – 49}} \, du$$

Express 0 as 98 -98

$$\int\limits {\frac{\sqrt{x + 49}}{x}} \, dx =\int\limits {\frac{2u^2-98+98}{u^2 – 49}} \, du$$

Split the fraction

$$\int{\frac{\sqrt{x + 49}}{x}} \, dx =\int( \frac{2u^2-98}{u^2 – 49}+\frac{98}{u^2 – 49} )\, du$$

$$\int{\frac{\sqrt{x + 49}}{x}} \, dx =\int( \frac{2(u^2-49)}{u^2 – 49}+\frac{98}{u^2 – 49} )\, du$$

$$\int{\frac{\sqrt{x + 49}}{x}} \, dx =\int( 2+\frac{98}{u^2 – 49} )\, du$$

Rewrite as:

$$\int{\frac{\sqrt{x + 49}}{x}} \, dx =\int( 2+\frac{98}{u^2 – 7^2} )\, du$$

Integrate

$$\int{\frac{\sqrt{x + 49}}{x}} \, dx =2u +\int\frac{98}{u^2 – 7^2} \, du$$

Remove constant (98)

$$\int{\frac{\sqrt{x + 49}}{x}} \, dx =2u +98\int\frac{1}{u^2 – 7^2} \, du$$

As a general rule,

$$\int \frac{1}{x^2 – a^2} \, dx = \frac{1}{2}\ln|\frac{x-a}{x+a}|$$

So, we have:

$$\int \frac{1}{u^2 – 7^2} \, du = \frac{1}{2}\ln|\frac{u-7}{u+7}|$$

$$\int{\frac{\sqrt{x + 49}}{x}} \, dx =2u +98\frac{1}{u^2 – 7^2} \, du$$ becomes

$$\int{\frac{\sqrt{x + 49}}{x}} \, dx =2u +98*\frac{1}{2}\ln|\frac{u-7}{u+7}|+c$$

$$\int{\frac{\sqrt{x + 49}}{x}} \, dx =2u +49\ln|\frac{u-7}{u+7}|+c$$

Substitute values for u

$$\int{\frac{\sqrt{x + 49}}{x}} \, dx =2\sqrt{x + 49} +49\ln|\frac{\sqrt{x + 49}-7}{\sqrt{x + 49}+7}|+c$$