What is the integral of x^3sqrt(a+bx^4) ?

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

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integral of x^3sqrt(a+bx^4)

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = a + b x^{4}

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

Simplify

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

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

Add a constant to the solution

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

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