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

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    2021-09-03T04:12:10+00:00

    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]

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