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

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

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integral of 2/(xsqrt(9x^2+1))

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

- ln 9 x^{2} + 1 + 1 + ln 9 x^{2} + 1 - 1 + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

= 2 \cdot \int \frac{1}{u ^{2} - 1} d u

Factor

u^{2} - 1 : - (- u^{2} + 1)

= 2 \cdot \int \frac{1}{- ( - u ^{2} + 1 )} d u

Take the constant out:

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

= 2 (- \int \frac{1}{- u ^{2} + 1} d u)

Use the common integral:

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= 2 (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = 9 x^{2} + 1

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

Simplify

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

= - ln 9 x^{2} + 1 + 1 + ln 9 x^{2} + 1 - 1

Add a constant to the solution

= - ln 9 x^{2} + 1 + 1 + ln 9 x^{2} + 1 - 1 + C

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

in v erse l a pl a ce \frac{4 s e ^{- 3 s}}{( s ^{2} + 4 )}

im pl i c i t \frac{d y}{d x}, y = 2 x - 5

n = 1 \sum \infty arctan (4 n)

in v erse l a pl a ce \frac{2 s ^{2} + 1}{s + 3}

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