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

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

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integral of t^2sqrt(t+8)

\int t^{2} t + 8 d t

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

Solution steps

\int t^{2} t + 8 d t

Apply u-substitution

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

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

Expand

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

= \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

= \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{2}{7} u^{\frac{7}{2}} - \frac{32}{5} u^{\frac{5}{2}} + \frac{128}{3} u^{\frac{3}{2}}

Substitute back

u = t + 8

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

Add a constant to the solution

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

y = \frac{2 x}{x - 1}

x \to 0 lim ((\frac{2 arccos ( x )}{π})^{\frac{1}{x}})

\int \frac{2 + x + x ^{2}}{x} d x

f (t) = \frac{sin ( t )}{t}

\int \frac{y}{e ^{- y}} d y

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