What is the integral of (2x-1)/(sqrt(x^2+9)) ?

The integral of (2x-1)/(sqrt(x^2+9)) is 2sqrt(x^2+9)-ln(1/3 |x+sqrt(9+x^2)|)+C

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integral of (2x-1)/(sqrt(x^2+9))

\int \frac{2 x - 1}{x ^{2} + 9} d x

2 x^{2} + 9 - ln (\frac{1}{3} x + 9 + x^{2}) + C

Solution steps

\int \frac{2 x - 1}{x ^{2} + 9} d x

Expand

\frac{2 x - 1}{x ^{2} + 9} : \frac{2 x}{x ^{2} + 9} - \frac{1}{x ^{2} + 9}

= \int \frac{2 x}{x ^{2} + 9} - \frac{1}{x ^{2} + 9} 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{2 x}{x ^{2} + 9} d x - \int \frac{1}{x ^{2} + 9} d x

\int \frac{2 x}{x ^{2} + 9} d x = 2 x^{2} + 9

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

= 2 x^{2} + 9 - ln (\frac{1}{3} x + 9 + x^{2})

Add a constant to the solution

= 2 x^{2} + 9 - ln (\frac{1}{3} x + 9 + x^{2}) + C

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What is the integral of (2x-1)/(sqrt(x^2+9)) ?
The integral of (2x-1)/(sqrt(x^2+9)) is 2sqrt(x^2+9)-ln(1/3 |x+sqrt(9+x^2)|)+C