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

The integral of (6x-5)/(3x^2+5) is ln|3x^2+5|-sqrt(5/3)arctan(sqrt(3/5)x)+C

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

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

ln 3 x^{2} + 5 - \frac{5}{3} arctan (\frac{3}{5} x) + C

Solution steps

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

Expand

\frac{6 x - 5}{3 x ^{2} + 5} : \frac{6 x}{3 x ^{2} + 5} - \frac{5}{3 x ^{2} + 5}

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

\int \frac{6 x}{3 x ^{2} + 5} d x = ln 3 x^{2} + 5

\int \frac{5}{3 x ^{2} + 5} d x = \frac{5}{3} arctan (\frac{3}{5} x)

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

Add a constant to the solution

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

y^{^{'}} = \frac{x ^{2} y ^{4} + 2 x ^{2}}{y ^{3}}

\int \frac{1 + e ^{2 x}}{e ^{x}} d x

y^{^{'}} = - 2 x y

\frac{d y}{d x} = cot (1 - 2 x^{2})

\int \frac{1}{z ^{3} z ^{2} - 64} d z

What is the integral of (6x-5)/(3x^2+5) ?
The integral of (6x-5)/(3x^2+5) is ln|3x^2+5|-sqrt(5/3)arctan(sqrt(3/5)x)+C