int (x+2)/(x^2+5x) dx
To solve using partial fraction method, the denominator of the integrand should be factored.
(x+2)/(x^2+5x) = (x + 2)/(x(x+5))
Then, express it as sum of fractions.
(x+2)/(x(x+5)) = A/x + B/(x +5)
To determine the values of A and B, multiply both sides by the LCD of the fractions present.
x(x+5)*(x+2)/(x(x+5)) = (A/x + B/(x +5))*x(x+5)
x+2=A(x+5)+Bx
Then, assign values to x in which either x or x+5 will become zero.
So, plug-in x=0 to get the value of A.
0+2=A(0+5)+B(0)
2=5A
2/5=A
Also, plug-in x=-5 to get the value of B.
-5+2=A(-5+5)+B(-5)
-3=A(0)+B(-5)
-3=-5B
3/5=B
So the partial fraction decomposition of the integrand is:
int (x+2)(x^2+5x)dx
=int (x+2)/(x(x+5))dx
= int (2/(5x) +3/(5(x+5)))dx
Then, express it as sum of two integrals.
= int 2/(5x)dx + int 3/(5(x+5))dx
= 2/5 int 1/xdx + 3/5 int 1/(x+5)dx
To take the integral of this, apply the formula int 1/u du = ln|u|+C .
=2/5ln|x| + 3/5ln|x+5| + C
Therefore, int (x+2)/(x^2+5x)==2/5ln|x| + 3/5ln|x+5| + C .
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