What is the integral of 2/((t-9)^2) ?

The integral of 2/((t-9)^2) is -2/(t-9)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 2/((t-9)^2)

\int \frac{2}{( t - 9 ) ^{2}} d t

- \frac{2}{t - 9} + C

Solution steps

\int \frac{2}{( t - 9 ) ^{2}} d t

Take the constant out:

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

= 2 \cdot \int \frac{1}{( t - 9 ) ^{2}} d t

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{u}

= 2 (- \frac{1}{u})

Substitute back

u = t - 9

= 2 (- \frac{1}{t - 9})

Simplify

2 (- \frac{1}{t - 9}) : - \frac{2}{t - 9}

= - \frac{2}{t - 9}

Add a constant to the solution

= - \frac{2}{t - 9} + C

\int 6 x^{2} + 2 x d x

\int x^{- 3} cos (\frac{1}{x}) d x

\int e^{\frac{1}{4 x}} d x

\int (u^{6} - 2 u^{5} - u^{3} + \frac{2}{7}) d u

\int 3 cos (5 θ) d θ