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

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

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integral of (2x^5+5x^4+2x^3-3/(x^2)+4)

\int (2 x^{5} + 5 x^{4} + 2 x^{3} - \frac{3}{x ^{2}} + 4) d x

\frac{x ^{6}}{3} + x^{5} + \frac{x ^{4}}{2} + \frac{3}{x} + 4 x + C

Solution steps

\int 2 x^{5} + 5 x^{4} + 2 x^{3} - \frac{3}{x ^{2}} + 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 2 x^{5} d x + \int 5 x^{4} d x + \int 2 x^{3} d x - \int \frac{3}{x ^{2}} d x + \int 4 d x

\int 2 x^{5} d x = \frac{x ^{6}}{3}

\int 5 x^{4} d x = x^{5}

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

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

\int 4 d x = 4 x

= \frac{x ^{6}}{3} + x^{5} + \frac{x ^{4}}{2} - (- \frac{3}{x}) + 4 x

Simplify

= \frac{x ^{6}}{3} + x^{5} + \frac{x ^{4}}{2} + \frac{3}{x} + 4 x

Add a constant to the solution

= \frac{x ^{6}}{3} + x^{5} + \frac{x ^{4}}{2} + \frac{3}{x} + 4 x + C

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