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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Expand

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

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

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

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

Substitute back

u = x + 1

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

Simplify

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

Add a constant to the solution

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

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