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

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

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

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

3 ln ∣ x ∣ + \frac{1}{x} - 2 ln ∣ x + 3 ∣ + C

Solution steps

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

Take the partial fraction of

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

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

\int \frac{3}{x} d x = 3 ln ∣ x ∣

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

\int \frac{2}{x + 3} d x = 2 ln ∣ x + 3 ∣

= 3 ln ∣ x ∣ - (- \frac{1}{x}) - 2 ln ∣ x + 3 ∣

Simplify

= 3 ln ∣ x ∣ + \frac{1}{x} - 2 ln ∣ x + 3 ∣

Add a constant to the solution

= 3 ln ∣ x ∣ + \frac{1}{x} - 2 ln ∣ x + 3 ∣ + C

x \to 2 lim (x^{2} + 3 x + 9)

\int 5 cos (6 x) d x

\int_{- π}^{π} cos (x) sin (2 x) d x

\int \frac{7}{x ^{2} x ^{2} + 4} d x

x \to 0 + lim (\frac{x}{14 x ^{4} - 3 x})

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