6DUsingPartialFractions

Integration Using Partial Fractions

Some integrals need to be solved by splitting a rational expression into partial fractions.

Example

math . \qquad \displaystyle {\int } \dfrac{8x+14}{(x-2)(x+4)} \; dx \qquad x \neq \lbrace 2,\; -4 \rbrace math

First split the expression into partial fractions

math . \qquad \text{Let } \; \dfrac{8x+14}{(x-2)(x+4)} = \dfrac{A}{x-2} + \dfrac{B}{x+4} math . math . \qquad \Rightarrow \; \; \dfrac{8x+14}{(x-2)(x+4)} = \dfrac{A(x+4)+B(x-2)}{(x-2)(x+4)} math . math . \qquad \Rightarrow \; \; 8x+14 = A(x+4) + B(x-2) math . math \\ . \qquad \text {Sub } \; x = -4 \; \Rightarrow \; -18=A(0)+B(-6) \; \Rightarrow \; B=3 \\. \\ . \qquad \text {Sub } \; x = 2 \; \Rightarrow \; 30=A(6)+B(0) \; \Rightarrow \; A=5 math . math . \qquad \text{Thus } \; \dfrac{8x+14}{(x-2)(x+4)} = \dfrac{5}{x-2} + \dfrac{3}{x+4} math . Hence

math \\ . \qquad \displaystyle {\int } \dfrac{8x+14}{(x-2)(x+4)} \; dx \\. \\ . \qquad = \displaystyle {\int } \dfrac{5}{x-2} + \dfrac{3}{x+4} \;dx \\. \\ . \qquad = 5\log_e \big| x-2 \big| +3\log_e \big| x+4 \big| + c math

Use log laws to tidy this up

math \\ . \qquad = \log_e \Big( \big| x-2 \big|^5 \Big) + \log_e \Big( \big| x+4 \big|^3 \Big) + c \\. \\ . \qquad = \log_e \Big( \big| x-2 \big|^5 \, \big| x+4 \big|^3 \Big) +c, \qquad x \neq \lbrace 2, \; -4 \rbrace math

NOTE: You will sometimes see this:

math . \qquad \text{Let } \; c = \log_e k \quad \text { integration constant } \qquad k > 0 math

So math \\ . \qquad \displaystyle {\int } \dfrac{8x+14}{(x-2)(x+4)} \;dx \\. \\ . \qquad = \log_e \Big( \big| x-2 \big|^5 \, \big| x+4 \big|^3 \Big) + \log_e k math . math . \qquad = \log_e \Big( k \, \big| x-2 \big|^5 \, \big| x+4 \big|^3 \Big) \qquad k>0 math

More Examples .