Math, asked by game07, 7 months ago

Evaluate f 3x+5 /dx
x²+3x-18​

Answers

Answered by cpkumar27
1

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

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