What is the integral of 1/(xsqrt(36x^2+1)) ?

The integral of 1/(xsqrt(36x^2+1)) is -1/2 ln|sqrt(36x^2+1)+1|+1/2 ln|sqrt(36x^2+1)-1|+C

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integral of 1/(xsqrt(36x^2+1))

\int \frac{1}{x 36 x ^{2} + 1} d x

- \frac{1}{2} ln 36 x^{2} + 1 + 1 + \frac{1}{2} ln 36 x^{2} + 1 - 1 + C

Solution steps

\int \frac{1}{x 36 x ^{2} + 1} d x

Apply u-substitution

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

= \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 = 36 x^{2} + 1

= - (\frac{ln 36 x ^{2} + 1 + 1}{2} - \frac{ln 36 x ^{2} + 1 - 1}{2})

Simplify

- (\frac{ln 36 x ^{2} + 1 + 1}{2} - \frac{ln 36 x ^{2} + 1 - 1}{2}) : - \frac{1}{2} ln 36 x^{2} + 1 + 1 + \frac{1}{2} ln 36 x^{2} + 1 - 1

= - \frac{1}{2} ln 36 x^{2} + 1 + 1 + \frac{1}{2} ln 36 x^{2} + 1 - 1

Add a constant to the solution

= - \frac{1}{2} ln 36 x^{2} + 1 + 1 + \frac{1}{2} ln 36 x^{2} + 1 - 1 + C

y^{^{'}} = - y^{\frac{1}{3}}

\frac{d}{d x} (2 sin (x) - x)

(- \frac{5}{2}, 1), (- \frac{1}{2}, 4)

\int \frac{5 x}{2} d x

\int \frac{32 x}{x ^{4} - 16} d x

What is the integral of 1/(xsqrt(36x^2+1)) ?
The integral of 1/(xsqrt(36x^2+1)) is -1/2 ln|sqrt(36x^2+1)+1|+1/2 ln|sqrt(36x^2+1)-1|+C