What is the integral from 1 to t of 1/((2x+1)^3) ?

The integral from 1 to t of 1/((2x+1)^3) is (t^2+t-2)/(9(2t+1)^2)

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integral from 1 to t of 1/((2x+1)^3)

\int_{1}^{t} \frac{1}{( 2 x + 1 ) ^{3}} d x

\frac{t ^{2} + t - 2}{9 ( 2 t + 1 ) ^{2}}

Solution steps

\int_{1}^{t} \frac{1}{( 2 x + 1 ) ^{3}} d x

Apply u-substitution

: \int_{3}^{1 + 2 t} \frac{1}{2 u ^{3}} d u

= \int_{3}^{1 + 2 t} \frac{1}{2 u ^{3}} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{3}^{1 + 2 t} \frac{1}{u ^{3}} d u

Apply the Power Rule

: [- \frac{1}{2 u ^{2}}]_{3}^{1 + 2 t}

= \frac{1}{2} [- \frac{1}{2 u ^{2}}]_{3}^{1 + 2 t}

Compute the boundaries:

- \frac{1}{2 ( 1 + 2 t ) ^{2}} + \frac{1}{18}

= \frac{1}{2} (- \frac{1}{2 ( 1 + 2 t ) ^{2}} + \frac{1}{18})

Simplify

= \frac{t ^{2} + t - 2}{9 ( 2 t + 1 ) ^{2}}

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What is the integral from 1 to t of 1/((2x+1)^3) ?
The integral from 1 to t of 1/((2x+1)^3) is (t^2+t-2)/(9(2t+1)^2)