What is the integral from 0 to 9 of (185)/((t+1)^3) ?

The integral from 0 to 9 of (185)/((t+1)^3) is 3663/40

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integral from 0 to 9 of (185)/((t+1)^3)

\int_{0}^{9} \frac{185}{( t + 1 ) ^{3}} d t

\frac{3663}{40}

Decimal

91.575

Solution steps

\int_{0}^{9} \frac{185}{( t + 1 ) ^{3}} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 185 \cdot \int_{0}^{9} \frac{1}{( t + 1 ) ^{3}} d t

Apply u-substitution

: \int_{1}^{10} \frac{1}{u ^{3}} d u

= 185 \cdot \int_{1}^{10} \frac{1}{u ^{3}} d u

Apply the Power Rule

: [- \frac{1}{2 u ^{2}}]_{1}^{10}

= 185 [- \frac{1}{2 u ^{2}}]_{1}^{10}

Compute the boundaries:

\frac{99}{200}

= 185 \cdot \frac{99}{200}

Simplify

= \frac{3663}{40}

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What is the integral from 0 to 9 of (185)/((t+1)^3) ?
The integral from 0 to 9 of (185)/((t+1)^3) is 3663/40