What is the integral of 8/(x^7)-6/(x^4)+3x ?

The integral of 8/(x^7)-6/(x^4)+3x is -4/(3x^6)+2/(x^3)+(3x^2)/2+C

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integral of 8/(x^7)-6/(x^4)+3x

\int \frac{8}{x ^{7}} - \frac{6}{x ^{4}} + 3 x d x

- \frac{4}{3 x ^{6}} + \frac{2}{x ^{3}} + \frac{3 x ^{2}}{2} + C

Solution steps

\int \frac{8}{x ^{7}} - \frac{6}{x ^{4}} + 3 x 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{8}{x ^{7}} d x - \int \frac{6}{x ^{4}} d x + \int 3 x d x

\int \frac{8}{x ^{7}} d x = - \frac{4}{3 x ^{6}}

\int \frac{6}{x ^{4}} d x = - \frac{2}{x ^{3}}

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

= - \frac{4}{3 x ^{6}} - (- \frac{2}{x ^{3}}) + \frac{3 x ^{2}}{2}

Simplify

= - \frac{4}{3 x ^{6}} + \frac{2}{x ^{3}} + \frac{3 x ^{2}}{2}

Add a constant to the solution

= - \frac{4}{3 x ^{6}} + \frac{2}{x ^{3}} + \frac{3 x ^{2}}{2} + C

y = 0.4 x^{1.2}

y = x (x - 8), (16, 32)

\int_{0}^{1} x^{2} - 2 x + 1 d x

f (x) = 3 x + 1, (5, 4)

\frac{d}{d x} (\frac{x ^{2} + 1}{x - 1})

What is the integral of 8/(x^7)-6/(x^4)+3x ?
The integral of 8/(x^7)-6/(x^4)+3x is -4/(3x^6)+2/(x^3)+(3x^2)/2+C