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

The integral of 2x^2sqrt(x^3+1) is 4/9 (x^3+1)^{3/2}+C

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

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

\frac{4}{9} (x^{3} + 1)^{\frac{3}{2}} + C

Solution steps

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

Take the constant out:

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

= 2 \cdot \int x^{2} x^{3} + 1 d x

Apply u-substitution

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

= 2 \cdot \int \frac{u}{3} d u

Take the constant out:

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

= 2 \cdot \frac{1}{3} \cdot \int u d u

Apply the Power Rule

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

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

Substitute back

u = x^{3} + 1

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

Simplify

2 \cdot \frac{1}{3} \cdot \frac{2}{3} (x^{3} + 1)^{\frac{3}{2}} : \frac{4}{9} (x^{3} + 1)^{\frac{3}{2}}

= \frac{4}{9} (x^{3} + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{4}{9} (x^{3} + 1)^{\frac{3}{2}} + C

x \to + 2 lim (\frac{2 x - 25}{x - 2})

f (t) = arcsin (\frac{1}{t})

x \to - 0.1 lim (\frac{sin ( 3 x )}{x})

x^{^{''}} + 3 x^{^{'}} + 6 x = 0

What is the integral of 2x^2sqrt(x^3+1) ?
The integral of 2x^2sqrt(x^3+1) is 4/9 (x^3+1)^{3/2}+C