Question

Express the integrand as a sum of partial fractions and evaluate the. Integral

Answer 1

The integral expression given is,

$\int \frac{x+3}{x^2+5x-6}dx$

Neglecting the integral sign, and solve the the polynomial as a partial fraction

$\frac{x+3}{x^2+5x-6}$

Let us factorize the denominator

$\begin{gathered} x^2+5x-6 \\ (x^2-x)+(6x-6) \\ ^{}x(x-1)+6(x-1) \\ (x+6)(x-1) \end{gathered}$

Therefore,

$\frac{x+3}{x^2+5x-6}=\frac{x+3}{(x+6)(x-1)}$

Expanding the expression using partial fraction

$\begin{gathered} \frac{x+3}{(x+6)(x-1)}=\frac{A}{x+6}+\frac{B}{x-1} \\ \frac{x+3}{(x+6)(x-1)}=\frac{A(x-1)+B(x+6)}{(x+6)(x-1)} \end{gathered}$

Multiply both sides by sides by (x+6)(x-1), we will have

$x+3=A(x-1)+B(x+6)$

Substiute x = 1, into the equation above and solve for the value of B

$\begin{gathered} 1+3=A(1-1)+B(1+6) \\ 4=A(0)+7B_{_{}} \\ 4=_{_{_{}}}0+7B \\ 4=7B \\ \frac{4}{7}=\frac{7B}{7} \\ \frac{4}{7}=B \\ \Rightarrow B=\frac{4}{7} \end{gathered}$

Substiute x = -6, into the equation above and solve for the value of A

$\begin{gathered} -6+3=A(-6-1)_{}+B(-6+6) \\ -3=A(-7)+B(0)=-7A+0 \\ -3=-7A \\ \frac{-3}{-7}=\frac{-7A}{-7} \\ \frac{3}{7}=A \\ \Rightarrow A=\frac{3}{7} \end{gathered}$

Hence,

$\begin{gathered} \frac{x+3}{(x+6)(x-1)}=\frac{3}{7(x+6)}+\frac{4}{7(x-1)} \\ \frac{x+3}{(x+6)(x-1)}=\frac{1}{7}\lbrack\frac{3}{(x+6)}+\frac{4}{(x-1)}\rbrack \end{gathered}$

Now let us now install our answer back into the integral form

$\int \frac{x+3}{x^2+5x-6}dx=\int \frac{1}{7}\lbrack\frac{3}{(x+6)}+\frac{4}{(x-1)}\rbrack dx$

Hence, the correct option is Option C.