What is the integral of 1/(49^2+9x-6) ?

The integral of 1/(49^2+9x-6) is 1/9 ln|9x+2395|+C

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integral of 1/(49^2+9x-6)

\int \frac{1}{4 9 ^{2} + 9 x - 6} d x

\frac{1}{9} ln ∣ 9 x + 2395 ∣ + C

Solution steps

\int \frac{1}{4 9 ^{2} + 9 x - 6} d x

Simplify

= \int \frac{1}{9 x + 2395} d x

Apply u-substitution

: \int \frac{1}{9 u} d u

= \int \frac{1}{9 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{9} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{9} ln ∣ u ∣

Substitute back

u = 9 x + 2395

= \frac{1}{9} ln ∣ 9 x + 2395 ∣

Add a constant to the solution

= \frac{1}{9} ln ∣ 9 x + 2395 ∣ + C

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What is the integral of 1/(49^2+9x-6) ?
The integral of 1/(49^2+9x-6) is 1/9 ln|9x+2395|+C