Question

Evaluate using partial fractions $\int \frac{7x^2+75x-150}{x^3-25x}dx$

• A. $5\ln |x|+8\ln |x-5|-7\ln |x+5|+c$
• B. $5\ln |x|+8\ln |x-5|-6\ln |x+5|+c$
• C. $6\ln |x|+8\ln |x-5|-7\ln |x+5|+c$
• D. $8\ln |x|+8\ln |x-5|-7\ln |x+5|+c$

Answer 1

The correct option is C $6\ln |x|+8\ln |x-5|-7\ln |x+5|+c$

Let $\frac{7x^2+75x-150}{x^3-25x}=\frac{A}{x}+\frac{B}{(x-5)}+\frac{C}{(x+5)}$ ..... (1)

Multiply $x^3-25x$ on both the sides, we get

$7x^2+75x-150=A(x-5)(x+5)+Bx.(x+5)+C.x(x-5)$

Put $x=0$ to find value of A

$\therefore -150=A(-5)\left(5\right)\Rightarrow A=6$

Put $x=5$ to find value of B

$\therefore 7(5{)}^2+75(5)-150=B(5\left)\right(10)\Rightarrow B=8$

Put $x=-5$ to find value of C

$\therefore 7(-5{)}^2+75(-5)-150=C(-5\left)\right(-10)\Rightarrow C=7$

Substitute the value of A,B and C in equation (1), we get

$\frac{7x^2+75x-150}{x^3-25x}=\frac{6}{x}+\frac{8}{(x-5)}+\frac{7}{(x+5)}$

Now, integrating on both sides, we get

$\int \frac{7x^2+75x-150}{x^3-25x}dx=\int \frac{6}{x}dx+\int \frac{8}{(x-5)}dx+\int \frac{7}{(x+5)}dx$

$=6\ln |x|+8\ln |x-5|+\ln |x+5|+c$