What is the integral of 1/(xsqrt(81x^2+1)) ?

The integral of 1/(xsqrt(81x^2+1)) is -1/2 ln|sqrt(81x^2+1)+1|+1/2 ln|sqrt(81x^2+1)-1|+C

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integral of 1/(xsqrt(81x^2+1))

\int \frac{1}{x 81 x ^{2} + 1} d x

- \frac{1}{2} ln 81 x^{2} + 1 + 1 + \frac{1}{2} ln 81 x^{2} + 1 - 1 + C

Solution steps

\int \frac{1}{x 81 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 = 81 x^{2} + 1

= - (\frac{ln 81 x ^{2} + 1 + 1}{2} - \frac{ln 81 x ^{2} + 1 - 1}{2})

Simplify

- (\frac{ln 81 x ^{2} + 1 + 1}{2} - \frac{ln 81 x ^{2} + 1 - 1}{2}) : - \frac{1}{2} ln 81 x^{2} + 1 + 1 + \frac{1}{2} ln 81 x^{2} + 1 - 1

= - \frac{1}{2} ln 81 x^{2} + 1 + 1 + \frac{1}{2} ln 81 x^{2} + 1 - 1

Add a constant to the solution

= - \frac{1}{2} ln 81 x^{2} + 1 + 1 + \frac{1}{2} ln 81 x^{2} + 1 - 1 + C

\frac{1}{x ^{5}}

\int - sec^{2} (x) + cos (2 x) d x

\frac{( 5 x - 2 )}{( x ^{2} + 1 )}

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

\frac{2 u ^{2}}{( u ^{2} + u ) ^{3}}

What is the integral of 1/(xsqrt(81x^2+1)) ?
The integral of 1/(xsqrt(81x^2+1)) is -1/2 ln|sqrt(81x^2+1)+1|+1/2 ln|sqrt(81x^2+1)-1|+C