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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 ( u - 2 ) u} d u

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

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}{( u - 2 ) u} d u

Complete the square

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

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

Apply u-substitution

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

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

Factor

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int x cos^{2} (5 x) d x

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