Find: \int \frac{3x+5}{x^2+3x-18}dx

 

 

 

 
 
 
 
 

Answers (1)

\int \frac{3x+5}{x^2+3x-18}dx

Let I =\int \frac{3x+5}{x^2+3x-18}dx

{3x+5} = A(2x +3) +B

Comparing coefficients of x

3 = 2A

\frac{3}{2} =A

Comparing constant terms

5 = 3A +B\Rightarrow 5 = \frac{9}{2} +B

B = \frac{1}{2}

\therefore I =\frac{3}{2} \int \frac{2x+3}{x^2+3x-18}dx + \frac{1}{2}\int \frac{dx}{x^2+3x-18}

Let I_1 =\frac{3}{2} \int \frac{2x+3}{x^2+3x-18}dx

put {x^2+3x-18} =t

(2x + 3)dx = dt

I_1 =\frac{3}{2} \int \frac{dt}{t}

I_1 =\frac{3}{2} \log|t| + c_1

I_1 =\frac{3}{2} \log|x^2+3x-18| + c_1\quad -(1)

Now, I_2 = \frac{1}{2}\int \frac{dx}{x^2+3x-18}

I_2 = \frac{1}{2}\int \frac{dx}{x^2+2x\cdot\frac{3}{2} + \left(\frac{3}{2} \right )^2 - \left(\frac{3}{2} \right )^2-18}

      = \frac{1}{2}\int \frac{dx}{\left(x + \frac{3}{2} \right )^2 - \left(\frac{9}{2} \right )^2}

      \Rightarrow \frac{1}{2}\times \frac{2}{2\times9}\log \left |\frac{x + \frac{3}{2}-\frac{9}{2}}{x +\frac{3}{2} + \frac{9}{2}}\right|+c_2

I_2 = \frac{1}{18}\log \left |\frac{x -3}{x +6}\right|+c_2

\therefore I = I_1 +I_2

I =\frac{3}{2} \log|x^2+3x-18| + \frac{1}{18}\log \left |\frac{x -3}{x +6}\right|+c

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