Question

Find: integration of 3x + 5/ x^2 + 3x -18​

Answer 1

Answer:

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Answer 2

$\int \frac{3x+5}{x^2+3x-18}dx=\frac{13}{9}ln\mid{x+6}\mid+\frac{14}{9}ln\mid{x-3}\mid$+C

Step-by-step explanation:

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

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

$\int \frac{3x+5}{(x^2+6x-3x-18)}dx$

$\int \frac{3x+5}{(x(x+6)-3(x+6)}dx$

$\int \frac{3x+5}{(x+6)(x-3)}dx$

Using partial fraction method

$\frac{3x+5}{(x+6)(x-3)}=\frac{A}{x+6}+\frac{B}{x-3}$

$\frac{3x+5}{(x+6)(x-3)}=\frac{A(x-3)+B(x+6)}{(x+6)(x-3)}$

$3x+5=A(x-3)+B(x+6)$...(1)

$x-3=0\implies x=3$

$x+6=0\implies x=-6$

Substitute x=3

$3(3)+5=0+B(3+6)$

$14=9B$

$B=\frac{14}{9}$

Substitute x=-6

$3(-6)+5=A(-6-3)+0$

$-18+5=-9A$

$-13=-9A$

$A=\frac{13}{9}$

Substitute the values

$\int \frac{3x+5}{x^2+3x-18}dx=\frac{13}{9}\int \frac{1}{x+6}dx+\frac{14}{9}\int\frac{1}{x-3}dx$

$\int \frac{3x+5}{x^2+3x-18}dx=\frac{13}{9}ln\mid{x+6}\mid+\frac{14}{9}ln\mid{x-3}\mid$+C

Using the formula:$\int\frac{1}{x}dx=ln x$

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