What is the integral from 0 to 1 of 1/((x+1)^5) ?

The integral from 0 to 1 of 1/((x+1)^5) is 15/64

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integral from 0 to 1 of 1/((x+1)^5)

\int_{0}^{1} \frac{1}{( x + 1 ) ^{5}} d x

\frac{15}{64}

Decimal

0.234375

Solution steps

\int_{0}^{1} \frac{1}{( x + 1 ) ^{5}} d x

Apply u-substitution

: \int_{1}^{2} \frac{1}{u ^{5}} d u

= \int_{1}^{2} \frac{1}{u ^{5}} d u

Apply the Power Rule

: [- \frac{1}{4 u ^{4}}]_{1}^{2}

= [- \frac{1}{4 u ^{4}}]_{1}^{2}

Compute the boundaries:

\frac{15}{64}

= \frac{15}{64}

\int_{- \infty}^{- 1} x^{- \frac{17}{7}} d x

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\int_{0}^{4} x^{2} - 9 d x

\int_{0}^{1} 4 x^{2} + 1 x d x

What is the integral from 0 to 1 of 1/((x+1)^5) ?
The integral from 0 to 1 of 1/((x+1)^5) is 15/64