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

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

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

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

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

Solution steps

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

Apply u-substitution

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

Factor

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

= \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

= - \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}

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

Substitute back

u = 9 x^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

\int (x^{8}) d x

t an g e n t f (x) = x^{3} - 9 x, (3, 0)

t an g e n t f (x) = \frac{4 x}{x + 2}, (6, 3)

d er i v a t i v e f (x) = x^{2} + 5 x

\frac{d}{d x} (e^{l n (x^{2} + 1) + x})

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