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

The integral of x/((x+t)^2) is ln|t+x|+t/(t+x)+C

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

\int \frac{x}{( x + t ) ^{2}}

ln ∣ t + x ∣ + \frac{t}{t + x} + C

Solution steps

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

Apply u-substitution

: \int \frac{u - t}{u ^{2}} d u

= \int \frac{u - t}{u ^{2}} d u

Expand

\frac{u - t}{u ^{2}} : \frac{1}{u} - \frac{t}{u ^{2}}

= \int \frac{1}{u} - \frac{t}{u ^{2}} 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

= \int \frac{1}{u} d u - \int \frac{t}{u ^{2}} d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{t}{u ^{2}} d u = - \frac{t}{u}

= ln ∣ u ∣ - (- \frac{t}{u})

Substitute back

u = t + x

= ln ∣ t + x ∣ - (- \frac{t}{t + x})

Simplify

= ln ∣ t + x ∣ + \frac{t}{t + x}

Add a constant to the solution

= ln ∣ t + x ∣ + \frac{t}{t + x} + C

\frac{d}{d x} (4 arcsin (\frac{2 x - 4}{3 x + 2}))

x \to 1 lim (\frac{3 - \frac{3}{x}}{x - 1})

e^{20}

y = \frac{12 x}{x ^{2} - 4}, (4, 4)

\frac{x + 2 x t + cos ( t )}{1 + t ^{2}}

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