What is the integral of 6xsqrt(x-5) ?

The integral of 6xsqrt(x-5) is 6(2/5 (x-5)^{5/2}+10/3 (x-5)^{3/2})+C

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integral of 6xsqrt(x-5)

\int 6 x x - 5 d x

6 (\frac{2}{5} (x - 5)^{\frac{5}{2}} + \frac{10}{3} (x - 5)^{\frac{3}{2}}) + C

Solution steps

\int 6 x x - 5 d x

Take the constant out:

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

= 6 \cdot \int x x - 5 d x

Apply u-substitution

: \int (u + 5) u d u

= 6 \cdot \int (u + 5) u d u

Expand

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

= 6 \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

= 6 (\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}}

= 6 (\frac{2}{5} u^{\frac{5}{2}} + \frac{10}{3} u^{\frac{3}{2}})

Substitute back

u = x - 5

= 6 (\frac{2}{5} (x - 5)^{\frac{5}{2}} + \frac{10}{3} (x - 5)^{\frac{3}{2}})

Add a constant to the solution

= 6 (\frac{2}{5} (x - 5)^{\frac{5}{2}} + \frac{10}{3} (x - 5)^{\frac{3}{2}}) + C

\int (5 x + 1) d x

\int 5 e^{- 5 t} d t

\int \frac{cos ^{3} ( t )}{sin ( t )} d t

\int x (3 x + 1) d x

\int x sin (\frac{1}{7}) x d x

What is the integral of 6xsqrt(x-5) ?
The integral of 6xsqrt(x-5) is 6(2/5 (x-5)^{5/2}+10/3 (x-5)^{3/2})+C