What is the integral of (x^4+3x^2+4)/((x+1)^2) ?

The integral of (x^4+3x^2+4)/((x+1)^2) is (x^3)/3-x^2+6x-10ln|x+1|-8/(x+1)+C

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integral of (x^4+3x^2+4)/((x+1)^2)

\int \frac{x ^{4} + 3 x ^{2} + 4}{( x + 1 ) ^{2}} d x

\frac{x ^{3}}{3} - x^{2} + 6 x - 10 ln ∣ x + 1 ∣ - \frac{8}{x + 1} + C

Solution steps

\int \frac{x ^{4} + 3 x ^{2} + 4}{( x + 1 ) ^{2}} d x

Long division

\frac{x ^{4} + 3 x ^{2} + 4}{( x + 1 ) ^{2}} : x^{2} - 2 x + 6 + \frac{- 10 x - 2}{x ^{2} + 2 x + 1}

= \int x^{2} - 2 x + 6 + \frac{- 10 x - 2}{x ^{2} + 2 x + 1} 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 x^{2} d x - \int 2 x d x + \int 6 d x + \int \frac{- 10 x - 2}{x ^{2} + 2 x + 1} d x

\int x^{2} d x = \frac{x ^{3}}{3}

\int 2 x d x = x^{2}

\int 6 d x = 6 x

\int \frac{- 10 x - 2}{x ^{2} + 2 x + 1} d x = - 10 ln ∣ x + 1 ∣ - \frac{8}{x + 1}

= \frac{x ^{3}}{3} - x^{2} + 6 x - 10 ln ∣ x + 1 ∣ - \frac{8}{x + 1}

Add a constant to the solution

= \frac{x ^{3}}{3} - x^{2} + 6 x - 10 ln ∣ x + 1 ∣ - \frac{8}{x + 1} + C

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What is the integral of (x^4+3x^2+4)/((x+1)^2) ?
The integral of (x^4+3x^2+4)/((x+1)^2) is (x^3)/3-x^2+6x-10ln|x+1|-8/(x+1)+C