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## Homework Statement

[tex]\int_0^1 \! \frac{x-4}{x^2+3x+2} dx[/tex]

## Homework Equations

## The Attempt at a Solution

Factoring out the denominator...

[tex]x^2+3x+2 = (x+1)(x+2)[/tex]

Breaking up the main fraction into the separate fractions. Since both are linear, only a single constant is needed on top.

[tex]\frac{A}{x+1}+\frac{B}{x+2}[/tex]

Fining the coefficients (A and B) by choosing x values that make the other coefficient go to zero.

[tex]x-4 = A(x+2) + B(x+1)[/tex]

[tex]x=-2[/tex]

[tex]-6=B(-1)[/tex]

[tex]6=B[/tex]

[tex]x=-1[/tex]

[tex]-5=A(1)[/tex]

[tex]-5=A[/tex]

[tex]x=-2[/tex]

[tex]-6=B(-1)[/tex]

[tex]6=B[/tex]

[tex]x=-1[/tex]

[tex]-5=A(1)[/tex]

[tex]-5=A[/tex]

Back in the integral they go. Also, breaking up the main integral into a sum of integrals.

[tex]\int \frac{-5}{x+1}dx + \int \frac{6}{x+2} dx[/tex]

[tex]-5ln(x+1) + 6ln(x+2) + c[/tex]

[tex]-5ln(x+1) + 6ln(x+2) + c[/tex]

Now using the fundamental theorem of calculus to find the value.

[tex](-5ln(1+1) + 6ln(1+2))-(-5ln(0+1) + 6ln(0+2))[/tex]

Result:

[tex]6ln(3)-11ln(2)[/tex]

Wrong. Aw.