What is the integral of xsqrt(25x-8) ?

The integral of xsqrt(25x-8) is 1/625 (2/5 (25x-8)^{5/2}+16/3 (25x-8)^{3/2})+C

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integral of xsqrt(25x-8)

\int x 25 x - 8 d x

\frac{1}{625} (\frac{2}{5} (25 x - 8)^{\frac{5}{2}} + \frac{16}{3} (25 x - 8)^{\frac{3}{2}}) + C

Solution steps

\int x 25 x - 8 d x

Apply u-substitution

: \int \frac{u ( u + 8 )}{625} d u

= \int \frac{u ( u + 8 )}{625} d u

Take the constant out:

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

= \frac{1}{625} \cdot \int u (u + 8) d u

Expand

u (u + 8) : u^{\frac{3}{2}} + 8 u

= \frac{1}{625} \cdot \int u^{\frac{3}{2}} + 8 u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{625} (\int u^{\frac{3}{2}} d u + \int 8 u d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

\int 8 u d u = \frac{16}{3} u^{\frac{3}{2}}

= \frac{1}{625} (\frac{2}{5} u^{\frac{5}{2}} + \frac{16}{3} u^{\frac{3}{2}})

Substitute back

u = 25 x - 8

= \frac{1}{625} (\frac{2}{5} (25 x - 8)^{\frac{5}{2}} + \frac{16}{3} (25 x - 8)^{\frac{3}{2}})

Add a constant to the solution

= \frac{1}{625} (\frac{2}{5} (25 x - 8)^{\frac{5}{2}} + \frac{16}{3} (25 x - 8)^{\frac{3}{2}}) + C

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What is the integral of xsqrt(25x-8) ?
The integral of xsqrt(25x-8) is 1/625 (2/5 (25x-8)^{5/2}+16/3 (25x-8)^{3/2})+C