What is the integral of sqrt(1+4/(x^2)) ?

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

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integral of sqrt(1+4/(x^2))

\int 1 + \frac{4}{x ^{2}} d x

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

Solution steps

\int 1 + \frac{4}{x ^{2}} d x

Join

1 + \frac{4}{x ^{2}} : \frac{x ^{2} + 4}{x ^{2}}

= \int \frac{x ^{2} + 4}{x ^{2}} d x

Apply Integral Substitution

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

Take the constant out:

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

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

Apply u-substitution

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

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

Factor

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

= 2 \cdot \int \frac{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

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

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

= 2 (- \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

= 2 (- (\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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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