What is the integral of ((2x)/(x^2-36)+1/x+7) ?

The integral of ((2x)/(x^2-36)+1/x+7) is ln|x^2-36|+ln|x|+7x+C

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integral of ((2x)/(x^2-36)+1/x+7)

\int (\frac{2 x}{x ^{2} - 36} + \frac{1}{x} + 7) d x

ln x^{2} - 36 + ln ∣ x ∣ + 7 x + C

Solution steps

\int \frac{2 x}{x ^{2} - 36} + \frac{1}{x} + 7 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{2 x}{x ^{2} - 36} d x + \int \frac{1}{x} d x + \int 7 d x

\int \frac{2 x}{x ^{2} - 36} d x = ln x^{2} - 36

\int \frac{1}{x} d x = ln ∣ x ∣

\int 7 d x = 7 x

= ln x^{2} - 36 + ln ∣ x ∣ + 7 x

Add a constant to the solution

= ln x^{2} - 36 + ln ∣ x ∣ + 7 x + C

What is the integral of ((2x)/(x^2-36)+1/x+7) ?
The integral of ((2x)/(x^2-36)+1/x+7) is ln|x^2-36|+ln|x|+7x+C