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General Int (x^2-32x-69)/((4x^2+17x+4)*(x^2+4x+5)) dx
- Thread starter problem1
- Start date
Hello, problem1!
\(\displaystyle \text{The numerator is: }\;x^2+4x+4 - 36x - 72 - 1 \;=\;(x+2)^2-36(x+2) - 1\)
\(\displaystyle \text{The denominator has: }\;4x^2+16x+16 + x + 2 - 14 \;=\;4(x+2)^2 + (x+2) - 14\)
. . . . . . . . . . . . . . . . \(\displaystyle \text{and: }\;x^2 + 4x + 4 + 1 \;=\;(x+2)^2+1\)
\(\displaystyle \text{The integral becomes: }\;\int^{\infty}_0\frac{(x+2)^2 - 36(x+2) - 1}{[4(x+2)^2+(x+2)-14][(x+2)^2+1]}\,dx\)
\(\displaystyle \text{Substitute }\,u \:= \:x+2\quad\Rightarrow\quad du \,=\,dx\)
. . \(\displaystyle \text{and we have: }\;\int^{\infty}_2\frac{u^2-36u - 1}{(4u^2+u-14)(u^2+1)}\,du \;=\;\int^{\infty}_2\frac{u^2-36u-1}{(u+2)(4u-7)(u^2+1)}\,du\)
Now apply Partial Fractions . . .
. . . I'll wait in the car . . .
\(\displaystyle \int^{\infty}_0 \frac{x^2-32x-69}{(4x^2+17x+4)(x^2+4x+5)}\,dx\)
\(\displaystyle \text{The numerator is: }\;x^2+4x+4 - 36x - 72 - 1 \;=\;(x+2)^2-36(x+2) - 1\)
\(\displaystyle \text{The denominator has: }\;4x^2+16x+16 + x + 2 - 14 \;=\;4(x+2)^2 + (x+2) - 14\)
. . . . . . . . . . . . . . . . \(\displaystyle \text{and: }\;x^2 + 4x + 4 + 1 \;=\;(x+2)^2+1\)
\(\displaystyle \text{The integral becomes: }\;\int^{\infty}_0\frac{(x+2)^2 - 36(x+2) - 1}{[4(x+2)^2+(x+2)-14][(x+2)^2+1]}\,dx\)
\(\displaystyle \text{Substitute }\,u \:= \:x+2\quad\Rightarrow\quad du \,=\,dx\)
. . \(\displaystyle \text{and we have: }\;\int^{\infty}_2\frac{u^2-36u - 1}{(4u^2+u-14)(u^2+1)}\,du \;=\;\int^{\infty}_2\frac{u^2-36u-1}{(u+2)(4u-7)(u^2+1)}\,du\)
Now apply Partial Fractions . . .
. . . I'll wait in the car . . .
D
Deleted member 4993
Guest
Soroban
Why couldn't it be :
Why couldn't it be :
However, I'll still wait in the car....\(\displaystyle \int^{\infty}_0 \frac{x^2-32x-69}{(4x^2+17x+4)(x^2+4x+5)}\,dx\)
\(\displaystyle = \int^{\infty}_0 \frac{x^2-32x-69}{(4x+1)(x+4)(x+4)(x+1)}\,dx\)
\(\displaystyle = \int^{\infty}_0 \frac{x^2-32x-69}{(4x+1)(x+4)^2(x+1)}\,dx\)
Dr. Flim-Flam
Junior Member
- Joined
- Oct 10, 2007
- Messages
- 108
Why not 2x/(x^2+4x+5) + 4/(x^2+4x+5) -4/(4x+1) - 1/(x+4).
Then the antiderivative is ln|x^2+4x+5| + 4arctan(x+2)- 4arctan(x+2) - ln|4x+1| -ln|x+4|, which
reduces to -ln(5) over the interval 0 to infinity.
Note: In the comment immediately above my comment the denominator doesn't factor.
x^2+4x+5 is prime.
Then the antiderivative is ln|x^2+4x+5| + 4arctan(x+2)- 4arctan(x+2) - ln|4x+1| -ln|x+4|, which
reduces to -ln(5) over the interval 0 to infinity.
Note: In the comment immediately above my comment the denominator doesn't factor.
x^2+4x+5 is prime.
D
Deleted member 4993
Guest
Dr. Flim-Flam said:Why not 2x/(x^2+4x+5) + 4/(x^2+4x+5) -4/(4x+1) - 1/(x+4).
Then the antiderivative is ln|x^2+4x+5| + 4arctan(x+2)- 4arctan(x+2) - ln(4x+1) -ln(x+4|, which
reduces to -ln(5) over the interval 0 to infinity.
Note: In the comment immediately above my comment the denominator doesn't factor.
x^2+4x+5 is prime.<--- Correct
I stand corrected - that was a mistake