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

The integral of (sqrt(1+x^2))/(-2x) is 1/4 (ln|sqrt(1+x^2)+1|-ln|sqrt(1+x^2)-1|-2sqrt(1+x^2))+C

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integral of (sqrt(1+x^2))/(-2x)

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

\frac{1}{4} (ln 1 + x^{2} + 1 - ln 1 + x^{2} - 1 - 2 1 + x^{2}) + C

Solution steps

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

Take the constant out:

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

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

Simplify

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

Apply u-substitution

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

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

Factor

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

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

Take the constant out:

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

= (- \frac{1}{2}) (- \int \frac{u ^{2}}{- u ^{2} + 1} d u)

\frac{u ^{2}}{- u ^{2} + 1} = \frac{1}{- u ^{2} + 1} - 1

= (- \frac{1}{2}) (- \int \frac{1}{- u ^{2} + 1} - 1 d u)

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= (- \frac{1}{2}) (- (\int \frac{1}{- u ^{2} + 1} d u - \int 1 d u))

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

\int 1 d u = u

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

Substitute back

u = 1 + x^{2}

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

Simplify

(- \frac{1}{2}) (- (\frac{ln 1 + x ^{2} + 1}{2} - \frac{ln 1 + x ^{2} - 1}{2} - 1 + x^{2})) : \frac{1}{4} (ln 1 + x^{2} + 1 - ln 1 + x^{2} - 1 - 2 1 + x^{2})

= \frac{1}{4} (ln 1 + x^{2} + 1 - ln 1 + x^{2} - 1 - 2 1 + x^{2})

Add a constant to the solution

= \frac{1}{4} (ln 1 + x^{2} + 1 - ln 1 + x^{2} - 1 - 2 1 + x^{2}) + C

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What is the integral of (sqrt(1+x^2))/(-2x) ?
The integral of (sqrt(1+x^2))/(-2x) is 1/4 (ln|sqrt(1+x^2)+1|-ln|sqrt(1+x^2)-1|-2sqrt(1+x^2))+C