What is the integral from 0 to 1 of (36)/((2x+1)^3) ?

The integral from 0 to 1 of (36)/((2x+1)^3) is 8

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integral from 0 to 1 of (36)/((2x+1)^3)

\int_{0}^{1} \frac{36}{( 2 x + 1 ) ^{3}} d x

8

Solution steps

\int_{0}^{1} \frac{36}{( 2 x + 1 ) ^{3}} d x

Take the constant out:

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

= 36 \cdot \int_{0}^{1} \frac{1}{( 2 x + 1 ) ^{3}} d x

Apply u-substitution

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

= 36 \cdot \int_{1}^{3} \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

= 36 \cdot \frac{1}{2} \cdot \int_{1}^{3} \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 36 \cdot \frac{1}{2} [- \frac{1}{2 u ^{2}}]_{1}^{3}

Simplify

36 \cdot \frac{1}{2} [- \frac{1}{2 u ^{2}}]_{1}^{3} : 18 [- \frac{1}{2 u ^{2}}]_{1}^{3}

= 18 [- \frac{1}{2 u ^{2}}]_{1}^{3}

Compute the boundaries:

\frac{4}{9}

= 18 \cdot \frac{4}{9}

Simplify

= 8

\int \frac{1}{x} + \frac{1}{x ^{2}} + \frac{1}{x ^{3}} d x

\int sin (2 x) sin (7 x) d x

\frac{d y}{d x} = \frac{sin ( x )}{y}, y (0) = 1

\frac{\partial}{\partial x} (y \cdot e^{- x})

(3 x^{3})^{^{'}}

What is the integral from 0 to 1 of (36)/((2x+1)^3) ?
The integral from 0 to 1 of (36)/((2x+1)^3) is 8