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

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

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integral of 1/(x(k+x))

integral

\frac{1}{x ( k + x )}

- \frac{1}{k} (ln \frac{2}{k} x + 2 - ln (\frac{2}{∣ k ∣} ∣ x ∣)) + C

Solution steps

\int \frac{1}{x ( k + x )} d x

Complete the square

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

= \int \frac{1}{( x + \frac{k}{2} ) ^{2} - \frac{k ^{2}}{4}} d x

Apply u-substitution

: \int \frac{4}{4 u ^{2} - k ^{2}} d u

= \int \frac{4}{4 u ^{2} - k ^{2}} d u

Take the constant out:

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

= 4 \cdot \int \frac{1}{4 u ^{2} - k ^{2}} d u

Factor

4 u^{2} - k^{2} : - (- 4 u^{2} + k^{2})

= 4 \cdot \int \frac{1}{- ( - 4 u ^{2} + k ^{2} )} d u

Take the constant out:

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

= 4 (- \int \frac{1}{- 4 u ^{2} + k ^{2}} d u)

Apply Integral Substitution

= 4 (- \int \frac{1}{2 k ( - v ^{2} + 1 )} d v)

Take the constant out:

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

= 4 (- \frac{1}{2 k} \cdot \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}

= 4 (- \frac{1}{2 k} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

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

Simplify

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

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

Add a constant to the solution

= - \frac{1}{k} (ln \frac{2}{k} x + 2 - ln (\frac{2}{∣ k ∣} ∣ x ∣)) + C

sec (x) \cdot tan (x) + 5

\frac{1}{( x ^{2} + 3 ) ^{3}}

\frac{x ^{3}}{6}

6 e^{3 x}

2 x^{2} - 3

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