What is the integral from 1 to 5 of (x+1)sqrt(2x-1) ?

The integral from 1 to 5 of (x+1)sqrt(2x-1) is 186/5

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integral from 1 to 5 of (x+1)sqrt(2x-1)

\int_{1}^{5} (x + 1) 2 x - 1 d x

\frac{186}{5}

Decimal Notation

37.2

Solution steps

\int_{1}^{5} (x + 1) 2 x - 1 d x

Apply Integration By Parts

: [\frac{1}{3} (x + 1) (2 x - 1)^{\frac{3}{2}} - \int \frac{1}{3} (2 x - 1)^{\frac{3}{2}} d x]_{1}^{5}

= [\frac{1}{3} (x + 1) (2 x - 1)^{\frac{3}{2}} - \int \frac{1}{3} (2 x - 1)^{\frac{3}{2}} d x]_{1}^{5}

\int \frac{1}{3} (2 x - 1)^{\frac{3}{2}} d x = \frac{1}{15} (2 x - 1)^{\frac{5}{2}}

= [\frac{1}{3} (x + 1) (2 x - 1)^{\frac{3}{2}} - \frac{1}{15} (2 x - 1)^{\frac{5}{2}}]_{1}^{5}

Simplify

[\frac{1}{3} (x + 1) (2 x - 1)^{\frac{3}{2}} - \frac{1}{15} (2 x - 1)^{\frac{5}{2}}]_{1}^{5} : [\frac{1}{5} (x + 2) (2 x - 1)^{\frac{3}{2}}]_{1}^{5}

= [\frac{1}{5} (x + 2) (2 x - 1)^{\frac{3}{2}}]_{1}^{5}

Compute the boundaries:

\frac{186}{5}

= \frac{186}{5}

\int \frac{4}{2 x + 3} d x

\frac{t - t}{t ^{\frac{1}{3}}}

\frac{\partial}{\partial y} (4 y^{3})

\int x arcsin (2) x^{2} d x

\frac{d y}{d x} = y^{\frac{2}{3}}

What is the integral from 1 to 5 of (x+1)sqrt(2x-1) ?
The integral from 1 to 5 of (x+1)sqrt(2x-1) is 186/5