What is the integral of x/(a+bx) ?

The integral of x/(a+bx) is 1/(b^2)(a+bx-aln|a+bx|)+C

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integral of x/(a+bx)

\int \frac{x}{a + b x} d x

\frac{1}{b ^{2}} (a + b x - a ln ∣ a + b x ∣) + C

Solution steps

\int \frac{x}{a + b x} d x

Apply u-substitution

: \int \frac{u - a}{b ^{2} u} d u

= \int \frac{u - a}{b ^{2} u} d u

Take the constant out:

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

= \frac{1}{b ^{2}} \cdot \int \frac{u - a}{u} d u

Expand

\frac{u - a}{u} : 1 - \frac{a}{u}

= \frac{1}{b ^{2}} \cdot \int 1 - \frac{a}{u} d u

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}{b ^{2}} (\int 1 d u - \int \frac{a}{u} d u)

\int 1 d u = u

\int \frac{a}{u} d u = a ln ∣ u ∣

= \frac{1}{b ^{2}} (u - a ln ∣ u ∣)

Substitute back

u = a + b x

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

Add a constant to the solution

= \frac{1}{b ^{2}} (a + b x - a ln ∣ a + b x ∣) + C

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\int \frac{25}{25 + e ^{x}} d x

x y^{^{'}} + y = - 3 x y^{2}, y (1) = - 4

What is the integral of x/(a+bx) ?
The integral of x/(a+bx) is 1/(b^2)(a+bx-aln|a+bx|)+C