What is the integral of (x/(x^2+6x+8))^2 ?

The integral of (x/(x^2+6x+8))^2 is -2ln|x+2|-1/(x+2)+2ln|x+4|-4/(x+4)+C

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integral of (x/(x^2+6x+8))^2

\int (\frac{x}{x ^{2} + 6 x + 8})^{2} d x

- 2 ln ∣ x + 2 ∣ - \frac{1}{x + 2} + 2 ln ∣ x + 4 ∣ - \frac{4}{x + 4} + C

Solution steps

\int (\frac{x}{x ^{2} + 6 x + 8})^{2} d x

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \int \frac{x ^{2}}{( x ^{2} + 6 x + 8 ) ^{2}} d x

Take the partial fraction of

\frac{x ^{2}}{( x ^{2} + 6 x + 8 ) ^{2}} : - \frac{2}{x + 2} + \frac{1}{( x + 2 ) ^{2}} + \frac{2}{x + 4} + \frac{4}{( x + 4 ) ^{2}}

= \int - \frac{2}{x + 2} + \frac{1}{( x + 2 ) ^{2}} + \frac{2}{x + 4} + \frac{4}{( x + 4 ) ^{2}} 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 + 2} d x + \int \frac{1}{( x + 2 ) ^{2}} d x + \int \frac{2}{x + 4} d x + \int \frac{4}{( x + 4 ) ^{2}} d x

\int \frac{2}{x + 2} d x = 2 ln ∣ x + 2 ∣

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

\int \frac{2}{x + 4} d x = 2 ln ∣ x + 4 ∣

\int \frac{4}{( x + 4 ) ^{2}} d x = - \frac{4}{x + 4}

= - 2 ln ∣ x + 2 ∣ - \frac{1}{x + 2} + 2 ln ∣ x + 4 ∣ - \frac{4}{x + 4}

Add a constant to the solution

= - 2 ln ∣ x + 2 ∣ - \frac{1}{x + 2} + 2 ln ∣ x + 4 ∣ - \frac{4}{x + 4} + C

What is the integral of (x/(x^2+6x+8))^2 ?
The integral of (x/(x^2+6x+8))^2 is -2ln|x+2|-1/(x+2)+2ln|x+4|-4/(x+4)+C