What is the integral of xsqrt(27x-5) ?

The integral of xsqrt(27x-5) is 1/729 (2/5 (27x-5)^{5/2}+10/3 (27x-5)^{3/2})+C

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integral of xsqrt(27x-5)

\int x 27 x - 5 d x

\frac{1}{729} (\frac{2}{5} (27 x - 5)^{\frac{5}{2}} + \frac{10}{3} (27 x - 5)^{\frac{3}{2}}) + C

Solution steps

\int x 27 x - 5 d x

Apply u-substitution

: \int \frac{u ( u + 5 )}{729} d u

= \int \frac{u ( u + 5 )}{729} d u

Take the constant out:

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

= \frac{1}{729} \cdot \int u (u + 5) d u

Expand

u (u + 5) : u^{\frac{3}{2}} + 5 u

= \frac{1}{729} \cdot \int u^{\frac{3}{2}} + 5 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}{729} (\int u^{\frac{3}{2}} d u + \int 5 u d u)

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

\int 5 u d u = \frac{10}{3} u^{\frac{3}{2}}

= \frac{1}{729} (\frac{2}{5} u^{\frac{5}{2}} + \frac{10}{3} u^{\frac{3}{2}})

Substitute back

u = 27 x - 5

= \frac{1}{729} (\frac{2}{5} (27 x - 5)^{\frac{5}{2}} + \frac{10}{3} (27 x - 5)^{\frac{3}{2}})

Add a constant to the solution

= \frac{1}{729} (\frac{2}{5} (27 x - 5)^{\frac{5}{2}} + \frac{10}{3} (27 x - 5)^{\frac{3}{2}}) + C

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What is the integral of xsqrt(27x-5) ?
The integral of xsqrt(27x-5) is 1/729 (2/5 (27x-5)^{5/2}+10/3 (27x-5)^{3/2})+C