What is the integral of (t^2+4)((t^3)/3+4t)^{1/2} ?

The integral of (t^2+4)((t^3)/3+4t)^{1/2} is 2/3 (1/3 (t^3+12t))^{3/2}+C

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integral of (t^2+4)((t^3)/3+4t)^{1/2}

\int (t^{2} + 4) (\frac{t ^{3}}{3} + 4 t)^{\frac{1}{2}} d t

\frac{2}{3} (\frac{1}{3} (t^{3} + 12 t))^{\frac{3}{2}} + C

Solution steps

\int (t^{2} + 4) (\frac{t ^{3}}{3} + 4 t)^{\frac{1}{2}} d t

Apply u-substitution

: \int u^{\frac{1}{2}} d u

= \int u^{\frac{1}{2}} d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = \frac{t ^{3}}{3} + 4 t

= \frac{2}{3} (\frac{t ^{3}}{3} + 4 t)^{\frac{3}{2}}

Simplify

\frac{2}{3} (\frac{t ^{3}}{3} + 4 t)^{\frac{3}{2}} : \frac{2}{3} (\frac{1}{3} (t^{3} + 12 t))^{\frac{3}{2}}

= \frac{2}{3} (\frac{1}{3} (t^{3} + 12 t))^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{3} (\frac{1}{3} (t^{3} + 12 t))^{\frac{3}{2}} + C

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What is the integral of (t^2+4)((t^3)/3+4t)^{1/2} ?
The integral of (t^2+4)((t^3)/3+4t)^{1/2} is 2/3 (1/3 (t^3+12t))^{3/2}+C