What is the integral of 1/(xsqrt(1+16x^2)) ?

The integral of 1/(xsqrt(1+16x^2)) is -1/2 ln|sqrt(1+16x^2)+1|+1/2 ln|sqrt(1+16x^2)-1|+C

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integral of 1/(xsqrt(1+16x^2))

\int \frac{1}{x 1 + 16 x ^{2}} d x

- \frac{1}{2} ln 1 + 16 x^{2} + 1 + \frac{1}{2} ln 1 + 16 x^{2} - 1 + C

Solution steps

\int \frac{1}{x 1 + 16 x ^{2}} 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 = 1 + 16 x^{2}

= - (\frac{ln 1 + 16 x ^{2} + 1}{2} - \frac{ln 1 + 16 x ^{2} - 1}{2})

Simplify

- (\frac{ln 1 + 16 x ^{2} + 1}{2} - \frac{ln 1 + 16 x ^{2} - 1}{2}) : - \frac{1}{2} ln 1 + 16 x^{2} + 1 + \frac{1}{2} ln 1 + 16 x^{2} - 1

= - \frac{1}{2} ln 1 + 16 x^{2} + 1 + \frac{1}{2} ln 1 + 16 x^{2} - 1

Add a constant to the solution

= - \frac{1}{2} ln 1 + 16 x^{2} + 1 + \frac{1}{2} ln 1 + 16 x^{2} - 1 + C

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What is the integral of 1/(xsqrt(1+16x^2)) ?
The integral of 1/(xsqrt(1+16x^2)) is -1/2 ln|sqrt(1+16x^2)+1|+1/2 ln|sqrt(1+16x^2)-1|+C