Integral: int[1 to 0][(x - a)/((x + 1)(3x + 1)]dx = 0

[Guest]

$\displaystyle \int_{1}^0\frac{x-\alpha}{(x+1)(3x+1)}dx=0$

Find the value of the constant $\displaystyle \alpha$

 

[pka]

Use partial fractions:
\(\displaystyle \L
\int\limits_1^0 {\frac{{x - a}}{{\left( {x + 3} \right)\left( {3x + 1} \right)}}dx} = \int\limits_1^0 {\left[ {\frac{{a + 3}}{{8\left( {x + 3} \right)}} - \frac{{1 + 3a}}{{8\left( {3x + 1} \right)}}} \right]} dx = 0\)

 

[galactus]

\(\displaystyle \L\\\int\frac{x-a}{(x+1)(3x+1)}dx\)

Using partial fractions:

\(\displaystyle \L\\\frac{A}{x+1}+\frac{B}{3x+1}=x-a\)

We eventually find that \(\displaystyle \L\\A=\frac{a+1}{2}\) and $\displaystyle B=\frac{-3a}{2}-\frac{1}{2}$

\(\displaystyle \L\\(\frac{a+1}{2})\int_{1}^{0}\frac{1}{x+1}+\frac{1}{2}(-1-3a)\int_{1}^{0}\frac{1}{3x+1}dx\)

Now integrate, set to 0 and solve for a.