What is the integral of x^4sqrt(ax^5-b) ?

The integral of x^4sqrt(ax^5-b) is (2(ax^5-b)^{3/2})/(15a)+C

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integral of x^4sqrt(ax^5-b)

\int x^{4} a x^{5} - b d x

\frac{2 ( a x ^{5} - b ) ^{\frac{3}{2}}}{15 a} + C

Solution steps

\int x^{4} a x^{5} - b d x

Apply u-substitution

: \int \frac{u}{5 a} d u

= \int \frac{u}{5 a} d u

Take the constant out:

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

= \frac{1}{5 a} \cdot \int u d u

Apply the Power Rule

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

= \frac{1}{5 a} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = a x^{5} - b

= \frac{1}{5 a} \cdot \frac{2}{3} (a x^{5} - b)^{\frac{3}{2}}

Simplify

\frac{1}{5 a} \cdot \frac{2}{3} (a x^{5} - b)^{\frac{3}{2}} : \frac{2 ( a x ^{5} - b ) ^{\frac{3}{2}}}{15 a}

= \frac{2 ( a x ^{5} - b ) ^{\frac{3}{2}}}{15 a}

Add a constant to the solution

= \frac{2 ( a x ^{5} - b ) ^{\frac{3}{2}}}{15 a} + C

\frac{\partial}{\partial x} (ln (x + 6 y + 9 z))

\int_{0}^{\frac{π}{2}} tan (\frac{x}{2}) d x

25 y^{^{''}} - 40 y^{^{'}} + 16 y = 0, y (0) = 4, y^{^{'}} (0) = - 5

\int ln (x^{\frac{1}{2}}) d x

\frac{s}{( s - 1 ) ^{2}}

What is the integral of x^4sqrt(ax^5-b) ?
The integral of x^4sqrt(ax^5-b) is (2(ax^5-b)^{3/2})/(15a)+C