What is the integral of x^2sqrt(6x+5) ?

The integral of x^2sqrt(6x+5) is 1/567 (27x^2-18x+10)(6x+5)^{3/2}+C

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integral of x^2sqrt(6x+5)

\int x^{2} 6 x + 5 d x

\frac{1}{567} (27 x^{2} - 18 x + 10) (6 x + 5)^{\frac{3}{2}} + C

Solution steps

\int x^{2} 6 x + 5 d x

Apply u-substitution

: \int \frac{u ( u - 5 ) ^{2}}{216} d u

= \int \frac{u ( u - 5 ) ^{2}}{216} d u

Take the constant out:

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

= \frac{1}{216} \cdot \int u (u - 5)^{2} d u

Expand

u (u - 5)^{2} : u^{\frac{5}{2}} - 10 u^{\frac{3}{2}} + 25 u

= \frac{1}{216} \cdot \int u^{\frac{5}{2}} - 10 u^{\frac{3}{2}} + 25 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}{216} (\int u^{\frac{5}{2}} d u - \int 10 u^{\frac{3}{2}} d u + \int 25 u d u)

\int u^{\frac{5}{2}} d u = \frac{2}{7} u^{\frac{7}{2}}

\int 10 u^{\frac{3}{2}} d u = 4 u^{\frac{5}{2}}

\int 25 u d u = \frac{50}{3} u^{\frac{3}{2}}

= \frac{1}{216} (\frac{2}{7} u^{\frac{7}{2}} - 4 u^{\frac{5}{2}} + \frac{50}{3} u^{\frac{3}{2}})

Substitute back

u = 6 x + 5

= \frac{1}{216} (\frac{2}{7} (6 x + 5)^{\frac{7}{2}} - 4 (6 x + 5)^{\frac{5}{2}} + \frac{50}{3} (6 x + 5)^{\frac{3}{2}})

Simplify

\frac{1}{216} (\frac{2}{7} (6 x + 5)^{\frac{7}{2}} - 4 (6 x + 5)^{\frac{5}{2}} + \frac{50}{3} (6 x + 5)^{\frac{3}{2}}) : \frac{1}{567} (27 x^{2} - 18 x + 10) (6 x + 5)^{\frac{3}{2}}

= \frac{1}{567} (27 x^{2} - 18 x + 10) (6 x + 5)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{567} (27 x^{2} - 18 x + 10) (6 x + 5)^{\frac{3}{2}} + C

f (x) = 3 a^{3 x}

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x \to 1 lim (\frac{- 3 x - 1}{x - 1})

\int (e^{x} + 2)^{2} e^{x} d x

What is the integral of x^2sqrt(6x+5) ?
The integral of x^2sqrt(6x+5) is 1/567 (27x^2-18x+10)(6x+5)^{3/2}+C