Question

Evaluate I= integral of (2x^2+3x+4)/(x^2+6x+10)dx pls hlp !!!!

Answer 1

first do long divison

Answer 2

\[\int\limits{\frac {(2x^{2}+3x+4)}{(x^2+6x+10)}}dx\]

Answer 3

easier should be that first take 2 common from numerator,,then brake it into denominator + (something)
that something has 1 degree less than denominator..
now many methods to do this..you may use partial differentiation..

Answer 4

long division you'll get
\[\int\limits \left(\frac{-9 x-16}{x^2+6 x+10}+2\right) \, dx\]

Answer 5

how to solve -9x-16 /x^2+6x+10

Answer 6

to get \[\frac{11}{x^2+6x+10} - \frac{9(2x+6)}{2(x^2+6x+10)}\]\

Answer 7

in my txt book we have a formula like \[\int \frac{(px^2+qx+r)}{(ax^2+bx+c)} =A(ax^2+bx+c)+B d/dx(ax^2+bx+c) +c \]\

Answer 8

consider integral of form \[\int \frac{px^2+qx+r)}{ax^2+bx+c)}\]\ then let px^2+qx+r =A(ax^2+bx+c)+B d/dx(ax^2+bx+c) +c

Answer 9

i substituted in this form but i dnt know how to find A B and C

Answer 10

i.e let \[\ 2x^2+3x+4 =A(x^2+6x+10) + B d/dx(x^2+6x+10)+c\]\

Answer 11

i.e \[\ 2x^2+3x+4=A(x^2+6x+10)+B (2x+6) +c\]\

Answer 12

if i substitute x=(-3) we get \[\ A+C=13\]\

Answer 13

what is the nxt step aftr this??? to find A ,B and C pls hlp

Answer 14

do you know trig sub?

Answer 15

u mean the formulas

Answer 16

\[\int\frac{dx}{a^2+x^2} = \frac {1}{a} tan^{-1} \frac{x}{a} +c\]\

Answer 17

\[\int\limits_{}^{}\frac{-9x-16}{x^2+6x+10} dx\]
\[\int\limits_{}^{}\frac{-9x-16}{(x^2+6x)+10} dx\]
\[\int\limits_{}^{}\frac{-9x-16}{(x^2+6x+9)+10-9} dx\]
\[\int\limits_{}^{}\frac{-9x-16}{(x+3)^2+1} dx\]

\[\tan(\theta)=\frac{x+3}{1}\]
\[\tan(\theta)=x+3\]
differentiating both sides
\[\sec^2(\theta) d \theta =1 dx\]
So we have
\[\int\limits_{}^{}\frac{-9(\tan(\theta)-3)-16}{\tan^2(\theta)+1} \sec^2(\theta) d \theta \]
\[\int\limits_{}^{}\frac{-9(\tan(\theta)-3)-16}{\sec^2(\theta)} \sec^2(\theta) d \theta \]
\[\int\limits_{}^{}[-9(\tan(\theta)-3)-16] d \theta \]
did i make a mistake
the rest should be easy

Answer 18

no not this type

Answer 19

my question s how to find the value of A B and C

Answer 20

pls hlp

Answer 21

sustituting x=0 we get \[\ 10A+6B+C=4\]\

Answer 22

pls hlp

Answer 23

sam! as u said long division method aftr long division wat we have to do

Answer 24

\[\int\limits \left(\frac{-9 x-16}{x^2+6 x+10}+2\right) \, dx\]
Separate
\[\int\limits \frac{-9 x-16}{x^2+6 x+10} \, dx+\int\limits 2 \, dx\]
Separate
\[\int\limits 2 \, dx+\int\limits \left(\frac{11}{x^2+6 x+10}-\frac{9 (2 x+6)}{2 \left(x^2+6 x+10\right)}\right) \, dx\]
Separatte
\[11\int\limits \frac{1}{x^2+6 x+10} \, dx-\frac{9}{2}\int\limits \frac{2 x+6}{x^2+6 x+10} \, dx+\int\limits 2 \, dx\]
Let
\[u=x^2+6 x+10~~ and ~~ u=2 x+6\]
\[11\int\limits \frac{1}{x^2+6 x+10} \, dx-\frac{9}{2}\int\limits \frac{1}{u} \, du+\int\limits 2 \, dx\]

Complete the square then integration by trig substitution
\[-\frac{9}{2}\int\limits\limits \frac{1}{u} \, du+11\int\limits\limits \frac{1}{(x+3)^2+1} \, dx+\int\limits\limits 2 \, dx\]

We'll call "s" as another variable ,
\[11 \tan ^{-1}(s)-\frac{9}{2}\int\limits \frac{1}{u} \, du+\int\limits 2 \, dx\]
Integrate all
\[11 \tan ^{-1}(s)-\frac{9 \ln (u)}{2}+2 x+c\]
Result
\[-\frac{9}{2} \ln \left(x^2+6 x+10\right)+2 x+11 \tan ^{-1}(x+3)+c\]

Answer 25

how did u seperate \[\frac{(-9x-16)}{(x^2+6x+10) }\]\

Answer 26

\[\frac{-9 x-16}{x^2+6 x+10}\]
You multiply 2 from numerator and denominator

Answer 27

how ?? i didnt understand

Answer 28

wait

Answer 29

(-16-9x)/(x^(2)+6x+10)

(11-9(x+3))/(x^(2)+6x+10)

(11)/(x^(2)+6x+10)-(9(x+3))/(x^(2)+6x+10)

(11)/(x^(2)+6x+10)-((9(x+3))/(x^(2)+6x+10))

(11)/(x^(2)+6x+10)-((18(x+3))/(2(x^(2)+6x+10)))

(11)/(x^(2)+6x+10)-(((18)(x+3))/(2(x^(2)+6x+10)))

(11)/(x^(2)+6x+10)-((9(2(x+3)))/(2(x^(2)+6x+10)))

(11)/(x^(2)+6x+10)-((9(2x+6))/(2(x^(2)+6x+10)))

Answer 30

@aafrin1410

Answer 31

\[\frac{-16-9x}{x^{2}+6x+10}\]
\[\frac{11-9(x+3)}{x^2+6x+10}\]
\[\frac{11}{x^2+6x+10}-\frac{9(x+3)}{x^2+6x+10}\]
\[\frac{11}{x^2+6 x+10}-\frac{9 (2 x+6)}{2 \left(x^2+6 x+10\right)}\]

Answer 32

thank you