What is the integral of (7x^5+x^3-3/(x^2)) ?

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

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integral of (7x^5+x^3-3/(x^2))

\int (7 x^{5} + x^{3} - \frac{3}{x ^{2}}) d x

\frac{7 x ^{6}}{6} + \frac{x ^{4}}{4} + \frac{3}{x} + C

Solution steps

\int 7 x^{5} + x^{3} - \frac{3}{x ^{2}} 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 7 x^{5} d x + \int x^{3} d x - \int \frac{3}{x ^{2}} d x

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

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

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

= \frac{7 x ^{6}}{6} + \frac{x ^{4}}{4} - (- \frac{3}{x})

Simplify

= \frac{7 x ^{6}}{6} + \frac{x ^{4}}{4} + \frac{3}{x}

Add a constant to the solution

= \frac{7 x ^{6}}{6} + \frac{x ^{4}}{4} + \frac{3}{x} + C

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