What is the integral of (x^2)/((x+1)^{50)} ?

The integral of (x^2)/((x+1)^{50)} is -1/(47(x+1)^{47)}+1/(24(x+1)^{48)}-1/(49(x+1)^{49)}+C

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integral of (x^2)/((x+1)^{50)}

\int \frac{x ^{2}}{( x + 1 ) ^{50}} d x

- \frac{1}{47 ( x + 1 ) ^{47}} + \frac{1}{24 ( x + 1 ) ^{48}} - \frac{1}{49 ( x + 1 ) ^{49}} + C

Solution steps

\int \frac{x ^{2}}{( x + 1 ) ^{50}} d x

Apply u-substitution

= \int \frac{( u - 1 ) ^{2}}{u ^{50}} d u

Expand

\frac{( u - 1 ) ^{2}}{u ^{50}} : \frac{1}{u ^{48}} - \frac{2}{u ^{49}} + \frac{1}{u ^{50}}

= \int \frac{1}{u ^{48}} - \frac{2}{u ^{49}} + \frac{1}{u ^{50}} d u

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{1}{u ^{48}} d u - \int \frac{2}{u ^{49}} d u + \int \frac{1}{u ^{50}} d u

\int \frac{1}{u ^{48}} d u = - \frac{1}{47 u ^{47}}

\int \frac{2}{u ^{49}} d u = - \frac{1}{24 u ^{48}}

\int \frac{1}{u ^{50}} d u = - \frac{1}{49 u ^{49}}

= - \frac{1}{47 u ^{47}} - (- \frac{1}{24 u ^{48}}) - \frac{1}{49 u ^{49}}

Substitute back

u = x + 1

= - \frac{1}{47 ( x + 1 ) ^{47}} - (- \frac{1}{24 ( x + 1 ) ^{48}}) - \frac{1}{49 ( x + 1 ) ^{49}}

Simplify

= - \frac{1}{47 ( x + 1 ) ^{47}} + \frac{1}{24 ( x + 1 ) ^{48}} - \frac{1}{49 ( x + 1 ) ^{49}}

Add a constant to the solution

= - \frac{1}{47 ( x + 1 ) ^{47}} + \frac{1}{24 ( x + 1 ) ^{48}} - \frac{1}{49 ( x + 1 ) ^{49}} + C

\int (2 x - 1) ln (17 x) d x

a re a \frac{2}{x}, 4 x, \frac{4}{x}

t a y l or 5 x

x y^{^{'}} - 4 y = x^{6} e^{x}

x \to \infty lim (\frac{x ^{99}}{e ^{x}})

What is the integral of (x^2)/((x+1)^{50)} ?
The integral of (x^2)/((x+1)^{50)} is -1/(47(x+1)^{47)}+1/(24(x+1)^{48)}-1/(49(x+1)^{49)}+C