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

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

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

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

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

Solution steps

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

Simplify

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

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

Expand

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

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

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

\int 4 d x = 4 x

\int \frac{5}{x ^{2}} d x = - \frac{5}{x}

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

Simplify

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

Add a constant to the solution

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

x^{2} + y^{2} = 23

\frac{d x}{d t} = 8 + 4 sin (\frac{t}{4}) - \frac{x}{100}

f (x) = \frac{4}{x}, at x = \frac{1}{4}

\int_{1}^{4} - 3 t ln (t) d t

\int_{0}^{5} 2 π x (\frac{1}{x + 4}) d x

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