What is the integral of x^3sqrt(ax^4+b) ?

The integral of x^3sqrt(ax^4+b) is ((ax^4+b)^{3/2})/(6a)+C

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integral of x^3sqrt(ax^4+b)

\int x^{3} a x^{4} + b d x

\frac{( a x ^{4} + b ) ^{\frac{3}{2}}}{6 a} + C

Solution steps

\int x^{3} a x^{4} + b d x

Apply u-substitution

: \int \frac{u}{4 a} d u

= \int \frac{u}{4 a} d u

Take the constant out:

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

= \frac{1}{4 a} \cdot \int u d u

Apply the Power Rule

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

= \frac{1}{4 a} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = a x^{4} + b

= \frac{1}{4 a} \cdot \frac{2}{3} (a x^{4} + b)^{\frac{3}{2}}

Simplify

\frac{1}{4 a} \cdot \frac{2}{3} (a x^{4} + b)^{\frac{3}{2}} : \frac{( a x ^{4} + b ) ^{\frac{3}{2}}}{6 a}

= \frac{( a x ^{4} + b ) ^{\frac{3}{2}}}{6 a}

Add a constant to the solution

= \frac{( a x ^{4} + b ) ^{\frac{3}{2}}}{6 a} + C

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What is the integral of x^3sqrt(ax^4+b) ?
The integral of x^3sqrt(ax^4+b) is ((ax^4+b)^{3/2})/(6a)+C