What is the integral of (6x^3+1/(5x^7)-4) ?

The integral of (6x^3+1/(5x^7)-4) is (3x^4)/2-1/(30x^6)-4x+C

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integral of (6x^3+1/(5x^7)-4)

\int (6 x^{3} + \frac{1}{5 x ^{7}} - 4) d x

\frac{3 x ^{4}}{2} - \frac{1}{30 x ^{6}} - 4 x + C

Solution steps

\int 6 x^{3} + \frac{1}{5 x ^{7}} - 4 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 6 x^{3} d x + \int \frac{1}{5 x ^{7}} d x - \int 4 d x

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

\int \frac{1}{5 x ^{7}} d x = - \frac{1}{30 x ^{6}}

\int 4 d x = 4 x

= \frac{3 x ^{4}}{2} - \frac{1}{30 x ^{6}} - 4 x

Add a constant to the solution

= \frac{3 x ^{4}}{2} - \frac{1}{30 x ^{6}} - 4 x + C

What is the integral of (6x^3+1/(5x^7)-4) ?
The integral of (6x^3+1/(5x^7)-4) is (3x^4)/2-1/(30x^6)-4x+C