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

The integral of 1/(u^2+xu) is 1/x (-ln((|u+x|}{|x|})+ln|\frac{u+x)/x-1|)+C

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integral of 1/(u^2+xu)

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

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

Solution steps

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

Factor

\frac{1}{u ^{2} + xu}

= \int \frac{1}{u ( u + x )} d u

Apply u-substitution

: \int \frac{1}{( v - x ) v} d v

= \int \frac{1}{( v - x ) v} d v

v - x = x (\frac{v}{x} - 1)

= \int \frac{1}{x ( \frac{v}{x} - 1 ) v} d v

Take the constant out:

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

= \frac{1}{x} \cdot \int \frac{1}{( \frac{v}{x} - 1 ) v} d v

Apply u-substitution

: \int \frac{1}{w ( w - 1 )} d w

= \frac{1}{x} \cdot \int \frac{1}{w ( w - 1 )} d w

Take the partial fraction of

\frac{1}{w ( w - 1 )} : - \frac{1}{w} + \frac{1}{w - 1}

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

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}{x} (- \int \frac{1}{w} d w + \int \frac{1}{w - 1} d w)

\int \frac{1}{w} d w = ln ∣ w ∣

\int \frac{1}{w - 1} d w = ln ∣ w - 1 ∣

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

Substitute back

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

Apply absolute rule:

\frac{a}{b} = \frac{∣ a ∣}{∣ b ∣}

\frac{u + x}{x} = \frac{∣ u + x ∣}{∣ x ∣}

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

Add a constant to the solution

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

\int_{0}^{1} 1 - 2 x - 3 x^{2} d x

\frac{d y}{d x} y = (4 x - 7)^{2}

(7, 5), (3, 2)

x \to 0 lim (\frac{1 - cos ( a x )}{x ^{2}})

\frac{d}{d x} (sin^{3} (\frac{x}{3}))

What is the integral of 1/(u^2+xu) ?
The integral of 1/(u^2+xu) is 1/x (-ln((|u+x|}{|x|})+ln|\frac{u+x)/x-1|)+C