What is the integral of (5x+1)/(6x^2+7x-3) ?

The integral of (5x+1)/(6x^2+7x-3) is 8/33 ln|3x-1|+13/22 ln|2x+3|+C

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integral of (5x+1)/(6x^2+7x-3)

\int \frac{5 x + 1}{6 x ^{2} + 7 x - 3} d x

\frac{8}{33} ln ∣ 3 x - 1 ∣ + \frac{13}{22} ln ∣ 2 x + 3 ∣ + C

Solution steps

\int \frac{5 x + 1}{6 x ^{2} + 7 x - 3} d x

Take the partial fraction of

\frac{5 x + 1}{6 x ^{2} + 7 x - 3} : \frac{8}{11 ( 3 x - 1 )} + \frac{13}{11 ( 2 x + 3 )}

= \int \frac{8}{11 ( 3 x - 1 )} + \frac{13}{11 ( 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{8}{11 ( 3 x - 1 )} d x + \int \frac{13}{11 ( 2 x + 3 )} d x

\int \frac{8}{11 ( 3 x - 1 )} d x = \frac{8}{33} ln ∣ 3 x - 1 ∣

\int \frac{13}{11 ( 2 x + 3 )} d x = \frac{13}{22} ln ∣ 2 x + 3 ∣

= \frac{8}{33} ln ∣ 3 x - 1 ∣ + \frac{13}{22} ln ∣ 2 x + 3 ∣

Add a constant to the solution

= \frac{8}{33} ln ∣ 3 x - 1 ∣ + \frac{13}{22} ln ∣ 2 x + 3 ∣ + C

\int sin (x) + 1 d x

\int \frac{1}{( 4 x ^{2} + 9 ) ^{\frac{3}{2}}} d x

n = 0 \sum \infty 1 + 2 + 3 + 4 + 5

\frac{d}{d x} (\frac{e ^{a r c t a n (x)}}{x ^{2} + 1})

\int (\frac{x}{2} + 1) d x

What is the integral of (5x+1)/(6x^2+7x-3) ?
The integral of (5x+1)/(6x^2+7x-3) is 8/33 ln|3x-1|+13/22 ln|2x+3|+C