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

The integral of (x^4-x^2+1/(4x^3)-1/(x^2)) is (x^5)/5-(x^3)/3-1/(8x^2)+1/x+C

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

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

\frac{x ^{5}}{5} - \frac{x ^{3}}{3} - \frac{1}{8 x ^{2}} + \frac{1}{x} + C

Solution steps

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

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

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

\int \frac{1}{4 x ^{3}} d x = - \frac{1}{8 x ^{2}}

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

= \frac{x ^{5}}{5} - \frac{x ^{3}}{3} - \frac{1}{8 x ^{2}} - (- \frac{1}{x})

Simplify

= \frac{x ^{5}}{5} - \frac{x ^{3}}{3} - \frac{1}{8 x ^{2}} + \frac{1}{x}

Add a constant to the solution

= \frac{x ^{5}}{5} - \frac{x ^{3}}{3} - \frac{1}{8 x ^{2}} + \frac{1}{x} + C

\frac{3}{( s ^{2} + 4 )}

27 n + 3 q^{2} n

y = \frac{4 x - 1}{2 x + 5}

\frac{d}{d x} (sin (x^{3}) cos (x^{2}))

\frac{d y}{d x} = x^{2} - y - 2, y (0) = - 1

What is the integral of (x^4-x^2+1/(4x^3)-1/(x^2)) ?
The integral of (x^4-x^2+1/(4x^3)-1/(x^2)) is (x^5)/5-(x^3)/3-1/(8x^2)+1/x+C