What is the integral of sqrt(1+5x^4)(20x^3) ?

The integral of sqrt(1+5x^4)(20x^3) is 2/3 (1+5x^4)^{3/2}+C

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integral of sqrt(1+5x^4)(20x^3)

\int 1 + 5 x^{4} (20 x^{3}) d x

\frac{2}{3} (1 + 5 x^{4})^{\frac{3}{2}} + C

Solution steps

\int 1 + 5 x^{4} \cdot 20 x^{3} d x

Take the constant out:

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

= 20 \cdot \int 1 + 5 x^{4} x^{3} d x

Apply u-substitution

: \int \frac{u}{20} d u

= 20 \cdot \int \frac{u}{20} d u

Take the constant out:

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

= 20 \cdot \frac{1}{20} \cdot \int u d u

Apply the Power Rule

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

= 20 \cdot \frac{1}{20} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 1 + 5 x^{4}

= 20 \cdot \frac{1}{20} \cdot \frac{2}{3} (1 + 5 x^{4})^{\frac{3}{2}}

Simplify

20 \cdot \frac{1}{20} \cdot \frac{2}{3} (1 + 5 x^{4})^{\frac{3}{2}} : \frac{2}{3} (1 + 5 x^{4})^{\frac{3}{2}}

= \frac{2}{3} (1 + 5 x^{4})^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{3} (1 + 5 x^{4})^{\frac{3}{2}} + C

n = 1 \sum \infty n (\frac{3}{2})^{n}

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

y = cos (\frac{π t}{2} - \frac{π}{3})

h \to 0 lim (\frac{1 - cos ( 7 h )}{h})

- 2 x^{2} + 2 x

What is the integral of sqrt(1+5x^4)(20x^3) ?
The integral of sqrt(1+5x^4)(20x^3) is 2/3 (1+5x^4)^{3/2}+C