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

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

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integral of 3/(x^4)+4/(x^2)-6/(x^3)

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

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

Solution steps

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

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

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

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

= - \frac{1}{x ^{3}} - \frac{4}{x} - (- \frac{3}{x ^{2}})

Simplify

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

Add a constant to the solution

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

2 t y^{^{'}} + (4 + 12 t^{3} - 120 t^{2}) y = 0

y = (5 t - 1) (6 t - 3)^{- 1}

f (x) = x + 4 e^{x}

\frac{d}{d t} (a cos (wt))

y = 2 x^{2} + 8 x + 10

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