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

The integral of (x^2)/((x+1)^4) is -(x^2)/(3(x+1)^3)+(-2x-1)/(3(x+1)^2)+C

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

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

- \frac{x ^{2}}{3 ( x + 1 ) ^{3}} + \frac{- 2 x - 1}{3 ( x + 1 ) ^{2}} + C

Solution steps

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

Apply Integration By Parts

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

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

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

= - \frac{x ^{2}}{3 ( x + 1 ) ^{3}} - (- \frac{- 2 x - 1}{3 ( x + 1 ) ^{2}})

Simplify

= - \frac{x ^{2}}{3 ( x + 1 ) ^{3}} + \frac{- 2 x - 1}{3 ( x + 1 ) ^{2}}

Add a constant to the solution

= - \frac{x ^{2}}{3 ( x + 1 ) ^{3}} + \frac{- 2 x - 1}{3 ( x + 1 ) ^{2}} + C

x \to 0 lim (x - \frac{4}{x ^{2}})

\frac{d}{d x} (arcsech (sin (x)))

\int e^{x} x^{4} d x

\frac{d}{d θ} (a^{2} \cdot sin^{2} (θ) + b^{2} \cdot cos^{2} (θ))

f (x) = sin^{11} (x)

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