What is the integral of 4x^3sqrt(x^4-7) ?

The integral of 4x^3sqrt(x^4-7) is 2/3 (x^4-7)^{3/2}+C

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integral of 4x^3sqrt(x^4-7)

\int 4 x^{3} x^{4} - 7 d x

\frac{2}{3} (x^{4} - 7)^{\frac{3}{2}} + C

Solution steps

\int 4 x^{3} x^{4} - 7 d x

Take the constant out:

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

= 4 \cdot \int x^{3} x^{4} - 7 d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = x^{4} - 7

= 4 \cdot \frac{1}{4} \cdot \frac{2}{3} (x^{4} - 7)^{\frac{3}{2}}

Simplify

4 \cdot \frac{1}{4} \cdot \frac{2}{3} (x^{4} - 7)^{\frac{3}{2}} : \frac{2}{3} (x^{4} - 7)^{\frac{3}{2}}

= \frac{2}{3} (x^{4} - 7)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{3} (x^{4} - 7)^{\frac{3}{2}} + C

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

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

\int 8 sin^{2} (t) d t

What is the integral of 4x^3sqrt(x^4-7) ?
The integral of 4x^3sqrt(x^4-7) is 2/3 (x^4-7)^{3/2}+C