What is the integral of (7-1/(x^6))^2(1/(x^7)) ?

The integral of (7-1/(x^6))^2(1/(x^7)) is -(49)/(6x^6)+7/(6x^{12)}-1/(18x^{18)}+C

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integral of (7-1/(x^6))^2(1/(x^7))

\int (7 - \frac{1}{x ^{6}})^{2} (\frac{1}{x ^{7}}) d x

- \frac{49}{6 x ^{6}} + \frac{7}{6 x ^{12}} - \frac{1}{18 x ^{18}} + C

Solution steps

\int (7 - \frac{1}{x ^{6}})^{2} \frac{1}{x ^{7}} d x

Simplify

(7 - \frac{1}{x ^{6}})^{2} \frac{1}{x ^{7}} : \frac{( 7 x ^{6} - 1 ) ^{2}}{x ^{19}}

= \int \frac{( 7 x ^{6} - 1 ) ^{2}}{x ^{19}} d x

Expand

\frac{( 7 x ^{6} - 1 ) ^{2}}{x ^{19}} : \frac{49}{x ^{7}} - \frac{14}{x ^{13}} + \frac{1}{x ^{19}}

= \int \frac{49}{x ^{7}} - \frac{14}{x ^{13}} + \frac{1}{x ^{19}} d x

Apply the Sum Rule:

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

= \int \frac{49}{x ^{7}} d x - \int \frac{14}{x ^{13}} d x + \int \frac{1}{x ^{19}} d x

\int \frac{49}{x ^{7}} d x = - \frac{49}{6 x ^{6}}

\int \frac{14}{x ^{13}} d x = - \frac{7}{6 x ^{12}}

\int \frac{1}{x ^{19}} d x = - \frac{1}{18 x ^{18}}

= - \frac{49}{6 x ^{6}} - (- \frac{7}{6 x ^{12}}) - \frac{1}{18 x ^{18}}

Simplify

= - \frac{49}{6 x ^{6}} + \frac{7}{6 x ^{12}} - \frac{1}{18 x ^{18}}

Add a constant to the solution

= - \frac{49}{6 x ^{6}} + \frac{7}{6 x ^{12}} - \frac{1}{18 x ^{18}} + C

f (t) = e^{t} - 1

\frac{d}{d x} (3 x - x)

f (x) = \frac{ln ( x )}{x ^{4}}

y = 2 sin (x) + sin (2 x), x = 0, x = π

\frac{d y}{d t} = y + 1 + 2 \cdot sin (t)

What is the integral of (7-1/(x^6))^2(1/(x^7)) ?
The integral of (7-1/(x^6))^2(1/(x^7)) is -(49)/(6x^6)+7/(6x^{12)}-1/(18x^{18)}+C