What is the integral of (sqrt(10+x^2))/x ?

The integral of (sqrt(10+x^2))/x is

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integral of (sqrt(10+x^2))/x

\int \frac{10 + x ^{2}}{x} d x

- 10 (\frac{1}{2} ln \frac{10 + x ^{2}}{10} + 1 - \frac{1}{2} ln \frac{10 + x ^{2}}{10} - 1 - \frac{1}{10} (10 + x^{2})) + C

Solution steps

\int \frac{10 + x ^{2}}{x} d x

Apply u-substitution

: \int \frac{u ^{2}}{u ^{2} - 10} d u

= \int \frac{u ^{2}}{u ^{2} - 10} d u

Factor

u^{2} - 10 : - (- u^{2} + 10)

= \int \frac{u ^{2}}{- ( - u ^{2} + 10 )} d u

Take the constant out:

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

= - \int \frac{u ^{2}}{- u ^{2} + 10} d u

Apply Integral Substitution

= - \int \frac{10 v ^{2}}{- v ^{2} + 1} d v

Take the constant out:

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

= - 10 \cdot \int \frac{v ^{2}}{- v ^{2} + 1} d v

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

= - 10 \cdot \int \frac{1}{- v ^{2} + 1} - 1 d v

Apply the Sum Rule:

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

= - 10 (\int \frac{1}{- v ^{2} + 1} d v - \int 1 d v)

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

\int 1 d v = v

= - 10 (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2} - v)

Substitute back

= - 10 \frac{ln \frac{10 + x ^{2}}{10} + 1}{2} - \frac{ln \frac{10 + x ^{2}}{10} - 1}{2} - \frac{10 + x ^{2}}{10}

Simplify

- 10 \frac{ln \frac{10 + x ^{2}}{10} + 1}{2} - \frac{ln \frac{10 + x ^{2}}{10} - 1}{2} - \frac{10 + x ^{2}}{10} : - 10 (\frac{1}{2} ln \frac{10 + x ^{2}}{10} + 1 - \frac{1}{2} ln \frac{10 + x ^{2}}{10} - 1 - \frac{1}{10} (10 + x^{2}))

= - 10 (\frac{1}{2} ln \frac{10 + x ^{2}}{10} + 1 - \frac{1}{2} ln \frac{10 + x ^{2}}{10} - 1 - \frac{1}{10} (10 + x^{2}))

Add a constant to the solution

= - 10 (\frac{1}{2} ln \frac{10 + x ^{2}}{10} + 1 - \frac{1}{2} ln \frac{10 + x ^{2}}{10} - 1 - \frac{1}{10} (10 + x^{2})) + C

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