What is the integral of (a+b)^{-2} ?

The integral of (a+b)^{-2} is -1/(b+a)+C

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integral of (a+b)^{-2}

integral

(a + b)^{- 2}

- \frac{1}{b + a} + C

Solution steps

\int (a + b)^{- 2} d a

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \int \frac{1}{( b + a ) ^{2}} d a

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 = b + a

= - \frac{1}{b + a}

Add a constant to the solution

= - \frac{1}{b + a} + C

\frac{x}{1 - x}

\frac{- x}{( 1 + x )}

(1 - t^{2})^{3}

\frac{5 x ^{4}}{x ^{5} + 9}

x \cdot lo g_{10} (\frac{1}{x})