What is the integral of sqrt(1+4x^4)(16x^3) ?

The integral of sqrt(1+4x^4)(16x^3) is 2/3 (4x^4+1)^{3/2}+C

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integral of sqrt(1+4x^4)(16x^3)

\int 1 + 4 x^{4} (16 x^{3}) d x

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

Solution steps

\int 1 + 4 x^{4} \cdot 16 x^{3} d x

Take the constant out:

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

= 16 \cdot \int 1 + 4 x^{4} x^{3} d x

Apply u-substitution

= 16 \cdot \int \frac{1 + 4 u}{4} d u

Take the constant out:

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

= 16 \cdot \frac{1}{4} \cdot \int 1 + 4 u d u

Apply u-substitution

= 16 \cdot \frac{1}{4} \cdot \int \frac{v}{4} d v

Take the constant out:

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

= 16 \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot \int v d v

Apply the Power Rule

= 16 \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{2}{3} v^{\frac{3}{2}}

Substitute back

= 16 \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{2}{3} (1 + 4 x^{4})^{\frac{3}{2}}

Simplify

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

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

Add a constant to the solution

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

What is the integral of sqrt(1+4x^4)(16x^3) ?
The integral of sqrt(1+4x^4)(16x^3) is 2/3 (4x^4+1)^{3/2}+C