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

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

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

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (ln (2 x - 4))

\int_{0}^{1} (\frac{1}{1 + x ^{2}}) d x

x \to 6 lim (4)

\int 3 x^{2} x^{3} + 1 d x

\int \frac{9 x ^{3}}{1 + x ^{2}} d x

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