What is the integral of (x^5)/(sqrt(a-bx^3)) ?

The integral of (x^5)/(sqrt(a-bx^3)) is 1/(3b^2)(2/3 (a-bx^3)^{3/2}-2asqrt(a-bx^3))+C

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integral of (x^5)/(sqrt(a-bx^3))

\int \frac{x ^{5}}{a - b x ^{3}} d x

\frac{1}{3 b ^{2}} (\frac{2}{3} (a - b x^{3})^{\frac{3}{2}} - 2 a a - b x^{3}) + C

Solution steps

\int \frac{x ^{5}}{a - b x ^{3}} d x

Apply u-substitution

: \int \frac{u - a}{3 b ^{2} u} d u

= \int \frac{u - a}{3 b ^{2} u} d u

Take the constant out:

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

= \frac{1}{3 b ^{2}} \cdot \int \frac{u - a}{u} d u

Expand

\frac{u - a}{u} : u - \frac{a}{u}

= \frac{1}{3 b ^{2}} \cdot \int u - \frac{a}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3 b ^{2}} (\int u d u - \int \frac{a}{u} d u)

\int u d u = \frac{2}{3} u^{\frac{3}{2}}

\int \frac{a}{u} d u = 2 a u

= \frac{1}{3 b ^{2}} (\frac{2}{3} u^{\frac{3}{2}} - 2 a u)

Substitute back

u = a - b x^{3}

= \frac{1}{3 b ^{2}} (\frac{2}{3} (a - b x^{3})^{\frac{3}{2}} - 2 a a - b x^{3})

Add a constant to the solution

= \frac{1}{3 b ^{2}} (\frac{2}{3} (a - b x^{3})^{\frac{3}{2}} - 2 a a - b x^{3}) + C

\frac{d}{d x} (\frac{x ^{5}}{4 - x ^{4}})

\frac{d}{d x} ((5 x)^{9 x})

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

f (x) = x^{4} + 3 x^{3}, at x = 1

n = 1 \sum \infty \frac{2}{5 ^{n - 1}}

What is the integral of (x^5)/(sqrt(a-bx^3)) ?
The integral of (x^5)/(sqrt(a-bx^3)) is 1/(3b^2)(2/3 (a-bx^3)^{3/2}-2asqrt(a-bx^3))+C