What is the integral of (sqrt(t^4+7t))(4t^3+7) ?

The integral of (sqrt(t^4+7t))(4t^3+7) is 2/3 (t^4+7t)^{3/2}+C

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integral of (sqrt(t^4+7t))(4t^3+7)

\int (t^{4} + 7 t) (4 t^{3} + 7) d t

\frac{2}{3} (t^{4} + 7 t)^{\frac{3}{2}} + C

Solution steps

\int t^{4} + 7 t (4 t^{3} + 7) d t

Apply u-substitution

: \int u d u

= \int u d u

Apply the Power Rule

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

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

Substitute back

u = t^{4} + 7 t

= \frac{2}{3} (t^{4} + 7 t)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{3} (t^{4} + 7 t)^{\frac{3}{2}} + C

x \to - 9 lim (\frac{x ^{2} - 1}{9 - x})

\frac{\partial}{\partial x} (y ln (x y^{2} p^{3}))

x \to 0 + lim (3 (x)^{\frac{x}{2}})

\int_{0}^{7} \frac{1}{49 - x ^{2}} d x

cos (sin (3 x))

What is the integral of (sqrt(t^4+7t))(4t^3+7) ?
The integral of (sqrt(t^4+7t))(4t^3+7) is 2/3 (t^4+7t)^{3/2}+C