What is the integral of x^6(x^7-3)^{3/4} ?

The integral of x^6(x^7-3)^{3/4} is 4/49 (x^7-3)^{7/4}+C

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integral of x^6(x^7-3)^{3/4}

\int x^{6} (x^{7} - 3)^{\frac{3}{4}} d x

\frac{4}{49} (x^{7} - 3)^{\frac{7}{4}} + C

Solution steps

\int x^{6} (x^{7} - 3)^{\frac{3}{4}} d x

Apply u-substitution

= \int \frac{u ^{\frac{3}{4}}}{7} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int u^{\frac{3}{4}} d u

Apply the Power Rule

= \frac{1}{7} \cdot \frac{4}{7} u^{\frac{7}{4}}

Substitute back

u = x^{7} - 3

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

Simplify

\frac{1}{7} \cdot \frac{4}{7} (x^{7} - 3)^{\frac{7}{4}} : \frac{4}{49} (x^{7} - 3)^{\frac{7}{4}}

= \frac{4}{49} (x^{7} - 3)^{\frac{7}{4}}

Add a constant to the solution

= \frac{4}{49} (x^{7} - 3)^{\frac{7}{4}} + C

x \to 0 + lim ((ln (\frac{1}{x}))^{x})

\frac{d}{d x} (10 cos (\frac{3}{10} x))

d er i v a t i v e f (x) = x^{2} + 8 x

x \to \infty lim (x^{- x})

\int_{0}^{2} x (x^{2} + 1)^{5} d x

What is the integral of x^6(x^7-3)^{3/4} ?
The integral of x^6(x^7-3)^{3/4} is 4/49 (x^7-3)^{7/4}+C