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

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

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

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

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

Solution steps

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

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

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

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

Add a constant to the solution

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

f (y) = 3 y^{2}

\frac{d}{d x} (2 x^{3} - 3 x^{2} - 72 x + 15)

\frac{d}{d x} (\frac{1}{4} ln (cos (4 x)))

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

- x^{2} + 10 x - 16, 2 x - 4

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