What is the integral from 0 to 7/2 of (2x+1)^{-1/3} ?

The integral from 0 to 7/2 of (2x+1)^{-1/3} is 9/4

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integral from 0 to 7/2 of (2x+1)^{-1/3}

\int_{0}^{\frac{7}{2}} (2 x + 1)^{- \frac{1}{3}} d x

\frac{9}{4}

Decimal

2.25

Solution steps

\int_{0}^{\frac{7}{2}} (2 x + 1)^{- \frac{1}{3}} d x

Apply Integration By Parts

: [\frac{x - \int - \frac{2 x}{3 ( 2 x + 1 ) ^{\frac{4}{3}}} d x ( 2 x + 1 ) ^{\frac{1}{3}}}{( 2 x + 1 ) ^{\frac{1}{3}}}]_{0}^{\frac{7}{2}}

= [\frac{x - \int - \frac{2 x}{3 ( 2 x + 1 ) ^{\frac{4}{3}}} d x ( 2 x + 1 ) ^{\frac{1}{3}}}{( 2 x + 1 ) ^{\frac{1}{3}}}]_{0}^{\frac{7}{2}}

\int - \frac{2 x}{3 ( 2 x + 1 ) ^{\frac{4}{3}}} d x = - \frac{2 x + 3}{4 ( 2 x + 1 ) ^{\frac{1}{3}}}

= [\frac{x - ( - \frac{2 x + 3}{4 ( 2 x + 1 ) ^{\frac{1}{3}}} ) ( 2 x + 1 ) ^{\frac{1}{3}}}{( 2 x + 1 ) ^{\frac{1}{3}}}]_{0}^{\frac{7}{2}}

Simplify

[\frac{x - ( - \frac{2 x + 3}{4 ( 2 x + 1 ) ^{\frac{1}{3}}} ) ( 2 x + 1 ) ^{\frac{1}{3}}}{( 2 x + 1 ) ^{\frac{1}{3}}}]_{0}^{\frac{7}{2}} : [\frac{3}{4} (2 x + 1)^{\frac{2}{3}}]_{0}^{\frac{7}{2}}

= [\frac{3}{4} (2 x + 1)^{\frac{2}{3}}]_{0}^{\frac{7}{2}}

Compute the boundaries:

\frac{9}{4}

= \frac{9}{4}

\frac{3 x + 2}{5 x - 1}

\frac{\partial}{\partial z} (x^{2} + z^{2})

y^{^{''}} + y = 3 sin (2 t) + t cos (2 t)

x \to - 1 lim (1 - x^{2})

\int \frac{2}{3} x^{\frac{3}{2}} - 4 sin (x) + C d x

What is the integral from 0 to 7/2 of (2x+1)^{-1/3} ?
The integral from 0 to 7/2 of (2x+1)^{-1/3} is 9/4