What is the integral from 0 to 2 of 1/((x+1)^4) ?

The integral from 0 to 2 of 1/((x+1)^4) is 26/81

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integral from 0 to 2 of 1/((x+1)^4)

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

\frac{26}{81}

Solution steps

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

Apply u-substitution

: \int_{1}^{3} \frac{1}{u ^{4}} d u

= \int_{1}^{3} \frac{1}{u ^{4}} d u

Apply the Power Rule

: [- \frac{1}{3 u ^{3}}]_{1}^{3}

= [- \frac{1}{3 u ^{3}}]_{1}^{3}

Compute the boundaries:

\frac{26}{81}

= \frac{26}{81}

f (x) = (arccot (x))^{5}

\int (x + 1) (x - 2) d x

f (x) = \frac{1 - 5 x}{4 + x}

\frac{\partial}{\partial x} (π^{7})

\frac{d}{d x} (2 sin^{2} (2 x))

What is the integral from 0 to 2 of 1/((x+1)^4) ?
The integral from 0 to 2 of 1/((x+1)^4) is 26/81