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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply u-substitution

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

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

Factor

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

= 2 \cdot \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

= 2 (- \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}

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

Substitute back

u = x^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

\frac{e ^{- s}}{s \cdot ( s + 5 ) ^{2}}

\frac{d}{d x} (\frac{ln ( 3 x )}{3 x})

\int \frac{16}{( x + 2 ) ( x ^{2} - 4 )} d x

\int e^{6 x} sin (6 x) d x

\frac{d y}{d x} - y^{2} = 0

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