What is the integral of 1/((x+t)^2) ?

The integral of 1/((x+t)^2) is -1/(t+x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/((x+t)^2)

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

- \frac{1}{t + x} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{2}} d u

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

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{u}

Substitute back

u = t + x

= - \frac{1}{t + x}

Add a constant to the solution

= - \frac{1}{t + x} + C

x \to 1 lim (3 x - 1)

\frac{\partial}{\partial x} (x^{3} + z x + 10)

\int \frac{1}{u + 1} d u

\frac{d}{d x} (\frac{3}{5 x ^{5}})

x \to 1 + lim (\frac{1}{ln ( x )})