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

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

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integral of ((3x^4+8x^3-5x^2+2)/(4x^2))

\int (\frac{3 x ^{4} + 8 x ^{3} - 5 x ^{2} + 2}{4 x ^{2}}) d x

\frac{1}{4} x^{3} + x^{2} - \frac{5}{4} x - \frac{1}{2 x} + C

Solution steps

\int \frac{3 x ^{4} + 8 x ^{3} - 5 x ^{2} + 2}{4 x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int \frac{3 x ^{4} + 8 x ^{3} - 5 x ^{2} + 2}{x ^{2}} d x

Expand

\frac{3 x ^{4} + 8 x ^{3} - 5 x ^{2} + 2}{x ^{2}} : 3 x^{2} + 8 x - 5 + \frac{2}{x ^{2}}

= \frac{1}{4} \cdot \int 3 x^{2} + 8 x - 5 + \frac{2}{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

= \frac{1}{4} (\int 3 x^{2} d x + \int 8 x d x - \int 5 d x + \int \frac{2}{x ^{2}} d x)

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

\int 8 x d x = 4 x^{2}

\int 5 d x = 5 x

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

= \frac{1}{4} (x^{3} + 4 x^{2} - 5 x - \frac{2}{x})

Simplify

\frac{1}{4} (x^{3} + 4 x^{2} - 5 x - \frac{2}{x}) : \frac{1}{4} x^{3} + x^{2} - \frac{5}{4} x - \frac{1}{2 x}

= \frac{1}{4} x^{3} + x^{2} - \frac{5}{4} x - \frac{1}{2 x}

Add a constant to the solution

= \frac{1}{4} x^{3} + x^{2} - \frac{5}{4} x - \frac{1}{2 x} + C

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

\int 3 e^{5 x - 2} cos (7 - 4 x) d x

\int y y + 4 d y

\int \frac{1}{x cos ^{2} ( x )} d x

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