Evaluate f 3x+5 /dx
x²+3x-18
Answers
x
2
+3x−18
3x+5
\int \frac{3x+5}{x^2+3x-18}dx∫
x
2
+3x−18
3x+5
dx
\int \frac{3x+5}{(x^2+6x-3x-18)}dx∫
(x
2
+6x−3x−18)
3x+5
dx
\int \frac{3x+5}{(x(x+6)-3(x+6)}dx∫
(x(x+6)−3(x+6)
3x+5
dx
\int \frac{3x+5}{(x+6)(x-3)}dx∫
(x+6)(x−3)
3x+5
dx
Using partial fraction method
\frac{3x+5}{(x+6)(x-3)}=\frac{A}{x+6}+\frac{B}{x-3}
(x+6)(x−3)
3x+5
=
x+6
A
+
x−3
B
\frac{3x+5}{(x+6)(x-3)}=\frac{A(x-3)+B(x+6)}{(x+6)(x-3)}
(x+6)(x−3)
3x+5
=
(x+6)(x−3)
A(x−3)+B(x+6)
3x+5=A(x-3)+B(x+6)3x+5=A(x−3)+B(x+6) ...(1)
x-3=0\implies x=3x−3=0⟹x=3
x+6=0\implies x=-6x+6=0⟹x=−6
Substitute x=3
3(3)+5=0+B(3+6)3(3)+5=0+B(3+6)
14=9B14=9B
B=\frac{14}{9}B=
9
14
Substitute x=-6
3(-6)+5=A(-6-3)+03(−6)+5=A(−6−3)+0
-18+5=-9A−18+5=−9A
-13=-9A−13=−9A
A=\frac{13}{9}A=
9
13
Substitute the values
\int \frac{3x+5}{x^2+3x-18}dx=\frac{13}{9}\int \frac{1}{x+6}dx+\frac{14}{9}\int\frac{1}{x-3}dx∫
x
2
+3x−18
3x+5
dx=
9
13
∫
x+6
1
dx+
9
14
∫
x−3
1
dx
\int \frac{3x+5}{x^2+3x-18}dx=\frac{13}{9}ln\mid{x+6}\mid+\frac{14}{9}ln\mid{x-3}\mid∫
x
2
+3x−18
3x+5
dx=
9
13
ln∣x+6∣+
9
14
ln∣x−3∣ +C
Using the formula:\int\frac{1}{x}dx=ln x∫
x
1
dx=lnx