What is the integral of x^3sqrt(9+x^4) ?

The integral of x^3sqrt(9+x^4) is 1/6 (9+x^4)^{3/2}+C

Popular >

integral of x^3sqrt(9+x^4)

\int x^{3} 9 + x^{4} d x

\frac{1}{6} (9 + x^{4})^{\frac{3}{2}} + C

Solution steps

\int x^{3} 9 + x^{4} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = 9 + x^{4}

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

Simplify

\frac{1}{4} \cdot \frac{2}{3} (9 + x^{4})^{\frac{3}{2}} : \frac{1}{6} (9 + x^{4})^{\frac{3}{2}}

= \frac{1}{6} (9 + x^{4})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{6} (9 + x^{4})^{\frac{3}{2}} + C

\int \frac{1}{3} x^{3} cos (x^{4} + 2) d x

\int (2 x + 1)^{- 3} d x

\int \frac{( a + b x ^{5} )}{6 a x + b x ^{6}} d x

\int \frac{7}{tan ( t ) ( csc ( t ) + 4 sin ( t ) )} d t

\int \frac{z}{e ^{4 z}} d z

What is the integral of x^3sqrt(9+x^4) ?
The integral of x^3sqrt(9+x^4) is 1/6 (9+x^4)^{3/2}+C