What is the integral of (2/(x^4)-3/(x^3)+1/(x^2)) ?

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

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integral of (2/(x^4)-3/(x^3)+1/(x^2))

\int (\frac{2}{x ^{4}} - \frac{3}{x ^{3}} + \frac{1}{x ^{2}}) d x

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

Solution steps

\int \frac{2}{x ^{4}} - \frac{3}{x ^{3}} + \frac{1}{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 \frac{2}{x ^{4}} d x - \int \frac{3}{x ^{3}} d x + \int \frac{1}{x ^{2}} d x

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

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

\int \frac{1}{x ^{2}} d x = - \frac{1}{x}

= - \frac{2}{3 x ^{3}} - (- \frac{3}{2 x ^{2}}) - \frac{1}{x}

Simplify

= - \frac{2}{3 x ^{3}} + \frac{3}{2 x ^{2}} - \frac{1}{x}

Add a constant to the solution

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

\int (6 - x) x d x

\int \frac{x ^{2} + x - 1}{x ^{3} - x} d x

\int 2 x^{2} - 2 x d x

\int \frac{9 x}{4 3 x ^{2} + 8} d x

\int x^{3} + 3 x^{2} + 2 x d x

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