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

The integral of (5x-3)/(x^2+5x+4) is -8/3 ln|x+1|+23/3 ln|x+4|+C

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

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

- \frac{8}{3} ln ∣ x + 1 ∣ + \frac{23}{3} ln ∣ x + 4 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{5 x - 3}{x ^{2} + 5 x + 4} : - \frac{8}{3 ( x + 1 )} + \frac{23}{3 ( x + 4 )}

= \int - \frac{8}{3 ( x + 1 )} + \frac{23}{3 ( x + 4 )} 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{8}{3 ( x + 1 )} d x + \int \frac{23}{3 ( x + 4 )} d x

\int \frac{8}{3 ( x + 1 )} d x = \frac{8}{3} ln ∣ x + 1 ∣

\int \frac{23}{3 ( x + 4 )} d x = \frac{23}{3} ln ∣ x + 4 ∣

= - \frac{8}{3} ln ∣ x + 1 ∣ + \frac{23}{3} ln ∣ x + 4 ∣

Add a constant to the solution

= - \frac{8}{3} ln ∣ x + 1 ∣ + \frac{23}{3} ln ∣ x + 4 ∣ + C

\int \frac{162}{x ^{3}} + 2 x d x

\int (1 - t) (8 + t^{2}) d t

\int \frac{3 ( ln ( x ) ) ^{2}}{x} d x

\int 2 x (x^{2} + 8)^{9} d x

\int \frac{t}{t} d t

What is the integral of (5x-3)/(x^2+5x+4) ?
The integral of (5x-3)/(x^2+5x+4) is -8/3 ln|x+1|+23/3 ln|x+4|+C