What is the integral of x^2sqrt(7x+8) ?

The integral of x^2sqrt(7x+8) is 1/343 (2/7 (7x+8)^{7/2}-32/5 (7x+8)^{5/2}+128/3 (7x+8)^{3/2})+C

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integral of x^2sqrt(7x+8)

\int x^{2} 7 x + 8 d x

\frac{1}{343} (\frac{2}{7} (7 x + 8)^{\frac{7}{2}} - \frac{32}{5} (7 x + 8)^{\frac{5}{2}} + \frac{128}{3} (7 x + 8)^{\frac{3}{2}}) + C

Solution steps

\int x^{2} 7 x + 8 d x

Apply u-substitution

: \int \frac{u ( u - 8 ) ^{2}}{343} d u

= \int \frac{u ( u - 8 ) ^{2}}{343} d u

Take the constant out:

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

= \frac{1}{343} \cdot \int u (u - 8)^{2} d u

Expand

u (u - 8)^{2} : u^{\frac{5}{2}} - 16 u^{\frac{3}{2}} + 64 u

= \frac{1}{343} \cdot \int u^{\frac{5}{2}} - 16 u^{\frac{3}{2}} + 64 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}{343} (\int u^{\frac{5}{2}} d u - \int 16 u^{\frac{3}{2}} d u + \int 64 u d u)

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

\int 16 u^{\frac{3}{2}} d u = \frac{32}{5} u^{\frac{5}{2}}

\int 64 u d u = \frac{128}{3} u^{\frac{3}{2}}

= \frac{1}{343} (\frac{2}{7} u^{\frac{7}{2}} - \frac{32}{5} u^{\frac{5}{2}} + \frac{128}{3} u^{\frac{3}{2}})

Substitute back

u = 7 x + 8

= \frac{1}{343} (\frac{2}{7} (7 x + 8)^{\frac{7}{2}} - \frac{32}{5} (7 x + 8)^{\frac{5}{2}} + \frac{128}{3} (7 x + 8)^{\frac{3}{2}})

Add a constant to the solution

= \frac{1}{343} (\frac{2}{7} (7 x + 8)^{\frac{7}{2}} - \frac{32}{5} (7 x + 8)^{\frac{5}{2}} + \frac{128}{3} (7 x + 8)^{\frac{3}{2}}) + C

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What is the integral of x^2sqrt(7x+8) ?
The integral of x^2sqrt(7x+8) is 1/343 (2/7 (7x+8)^{7/2}-32/5 (7x+8)^{5/2}+128/3 (7x+8)^{3/2})+C