What is the integral of t^2sqrt(t^3+1) ?

The integral of t^2sqrt(t^3+1) is 2/9 (t^3+1)^{3/2}+C

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integral of t^2sqrt(t^3+1)

\int t^{2} t^{3} + 1 d t

\frac{2}{9} (t^{3} + 1)^{\frac{3}{2}} + C

Solution steps

\int t^{2} t^{3} + 1 d t

Apply u-substitution

: \int \frac{u}{3} d u

= \int \frac{u}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int u d u

Apply the Power Rule

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

= \frac{1}{3} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = t^{3} + 1

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

Simplify

\frac{1}{3} \cdot \frac{2}{3} (t^{3} + 1)^{\frac{3}{2}} : \frac{2}{9} (t^{3} + 1)^{\frac{3}{2}}

= \frac{2}{9} (t^{3} + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{9} (t^{3} + 1)^{\frac{3}{2}} + C

\int \frac{r ^{3}}{4 + r ^{2}} d r

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

x y^{^{'}} - 2 y = 12 x^{2}

\frac{d}{d x} (x^{2} + 5 x - 8)

- 18 y^{^{''}} - 14 y^{^{'}} = 0, y (3) = - 3, y^{^{'}} (3) = - 3

What is the integral of t^2sqrt(t^3+1) ?
The integral of t^2sqrt(t^3+1) is 2/9 (t^3+1)^{3/2}+C