What is the integral of xsqrt(9x+8) ?

The integral of xsqrt(9x+8) is 1/81 (2/5 (9x+8)^{5/2}-16/3 (9x+8)^{3/2})+C

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integral of xsqrt(9x+8)

\int x 9 x + 8 d x

\frac{1}{81} (\frac{2}{5} (9 x + 8)^{\frac{5}{2}} - \frac{16}{3} (9 x + 8)^{\frac{3}{2}}) + C

Solution steps

\int x 9 x + 8 d x

Apply u-substitution

: \int \frac{u ( u - 8 )}{81} d u

= \int \frac{u ( u - 8 )}{81} d u

Take the constant out:

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

= \frac{1}{81} \cdot \int u (u - 8) d u

Expand

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

= \frac{1}{81} \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}{81} (\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}{81} (\frac{2}{5} u^{\frac{5}{2}} - \frac{16}{3} u^{\frac{3}{2}})

Substitute back

u = 9 x + 8

= \frac{1}{81} (\frac{2}{5} (9 x + 8)^{\frac{5}{2}} - \frac{16}{3} (9 x + 8)^{\frac{3}{2}})

Add a constant to the solution

= \frac{1}{81} (\frac{2}{5} (9 x + 8)^{\frac{5}{2}} - \frac{16}{3} (9 x + 8)^{\frac{3}{2}}) + C

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What is the integral of xsqrt(9x+8) ?
The integral of xsqrt(9x+8) is 1/81 (2/5 (9x+8)^{5/2}-16/3 (9x+8)^{3/2})+C