Use partial fractions to rewrite the integral \int \frac{2x + 13}{x^2 + 8x + 15} dx as the sum of...

Question:

Use partial fractions to rewrite the integral {eq}\int \frac{2x + 13}{x^2 + 8x + 15} dx {/eq} as the sum of two integrals.

Partial Fraction:

A partial fraction is a method of splitting the complicated fraction into one or more fractions having a simpler denominator.

Here we have to use partial fractions to rewrite the given integral as sum of two integral. For this, first, we factor the denominator by using the concept of splitting the middle term and then we apply the partial fraction method.

We have to use partial fractions to rewrite the integral {eq}\int \dfrac{2x + 13}{x^2 + 8x + 15} dx {/eq} as the sum of two integrals.

Now:

{eq}\begin{align} \int \dfrac{2x + 13}{x^2 + 8x + 15} dx &=\int \dfrac{2x + 13}{x^2 + 5x+3x + 15} dx\\ &=\int \dfrac{2x + 13}{x(x+5)+3(x+5)} dx\\ &=\int \dfrac{2x + 13}{(x+3)(x+5)} dx \end{align} {/eq}

Applying the partial fraction to split the fraction into a simpler form:

{eq}\begin{align} \dfrac{2x + 13}{(x+3)(x+5)} &=\dfrac{A}{x+3}+\dfrac{B}{x+5}\\ 2x+13 &=A(x+5)+B(x+3) \end{align} {/eq}

Let {eq}x+5=0 \Rightarrow x=-5 {/eq}

{eq}\begin{align} 2(-5)+13 &=A(0)+B(-5+3)\\ -10+13 &=B(-2)\\ 3 &=-2B\\ B &=\dfrac{-3}{2} \end{align} {/eq}

Let {eq}x+3=0 \Rightarrow x=-3 {/eq}

{eq}\begin{align} 2(-3)+13 &=A(-3+5)+B(0)\\ -6+13 &=A(2)\\ 7 &=2A\\ A &=\dfrac{7}{2} \end{align} {/eq}

Applying these values, we have:

{eq}\begin{align} \dfrac{2x + 13}{(x+3)(x+5)} &=\dfrac{\dfrac{7}{2}}{x+3}+\dfrac{\dfrac{-3}{2}}{x+5}\\ \dfrac{2x + 13}{x^2 + 8x + 15} &=\dfrac{7}{2(x+3)}-\dfrac{3}{2(x+5)} \end{align} {/eq}

Therefore:

{eq}\color{blue}{\boxed{\int \dfrac{2x + 13}{x^2 + 8x + 15} dx=\int \dfrac{7}{2(x+3)}dx+\int \dfrac{-3}{2(x+5)}dx}} {/eq}