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

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

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

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

\frac{1}{2} ln x^{2} + 1 + 4 ln (x) + \frac{3}{x} + C

Solution steps

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

Expand

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

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

\int \frac{5 x ^{3}}{x ^{4} + x ^{2}} d x = \frac{5}{2} ln x^{2} + 1

\int \frac{3 x ^{2}}{x ^{4} + x ^{2}} d x = 3 arctan (x)

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

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

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

Simplify

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

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

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 1 + 4 ln (x) + \frac{3}{x} + C

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\int (5 x + 6)^{4} d x

\int y - 2 d y

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