What is the integral of (x^6)/(x^7-2) ?

The integral of (x^6)/(x^7-2) is 1/7 ln|x^7-2|+C

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integral of (x^6)/(x^7-2)

\int \frac{x ^{6}}{x ^{7} - 2} d x

\frac{1}{7} ln x^{7} - 2 + C

Solution steps

\int \frac{x ^{6}}{x ^{7} - 2} d x

Apply u-substitution

: \int \frac{1}{7 u} d u

= \int \frac{1}{7 u} 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 \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{7} ln ∣ u ∣

Substitute back

u = x^{7} - 2

= \frac{1}{7} ln x^{7} - 2

Add a constant to the solution

= \frac{1}{7} ln x^{7} - 2 + C

\int_{0}^{2} ((- x + 5) - (x^{2} - 9)) d x

\frac{d}{d x} (\frac{2 a}{x ^{3}})

\int \frac{12}{( x ^{2} - 9 )} d x

\int \frac{w ^{5}}{8 w ^{2} + 1} d w

f (x) = 3 x - 2 x^{2}, (- 2, - 14)

What is the integral of (x^6)/(x^7-2) ?
The integral of (x^6)/(x^7-2) is 1/7 ln|x^7-2|+C