What is the integral from b to a of (x-y)/(x+y) ?

The integral from b to a of (x-y)/(x+y) is 2xln|a+x|-2xln|b+x|+b-a

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integral from b to a of (x-y)/(x+y)

\int_{b}^{a} \frac{x - y}{x + y} d y

2 x ln ∣ a + x ∣ - 2 x ln ∣ b + x ∣ + b - a

Solution steps

\int_{b}^{a} \frac{x - y}{x + y} d y

Expand

\frac{x - y}{x + y} : \frac{x}{x + y} - \frac{y}{x + y}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{b}^{a} \frac{x}{x + y} d y - \int_{b}^{a} \frac{y}{x + y} d y

\int_{b}^{a} \frac{x}{x + y} d y = x (ln ∣ x + a ∣ - ln ∣ x + b ∣)

\int_{b}^{a} \frac{y}{x + y} d y = - x (ln ∣ x + a ∣ - ln ∣ x + b ∣) + a - b

= x (ln ∣ x + a ∣ - ln ∣ x + b ∣) - (- x (ln ∣ x + a ∣ - ln ∣ x + b ∣) + a - b)

Simplify

x (ln ∣ x + a ∣ - ln ∣ x + b ∣) - (- x (ln ∣ x + a ∣ - ln ∣ x + b ∣) + a - b) : 2 x ln ∣ a + x ∣ - 2 x ln ∣ b + x ∣ + b - a

= 2 x ln ∣ a + x ∣ - 2 x ln ∣ b + x ∣ + b - a

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\int_{- \frac{π}{4}}^{0} (tan (x))^{2} d x

What is the integral from b to a of (x-y)/(x+y) ?
The integral from b to a of (x-y)/(x+y) is 2xln|a+x|-2xln|b+x|+b-a