MHB Complex Integration: Solving $\int_0^1\frac{2t+i}{t^2+it^2+1}dt$

[Dustinsfl]

$\displaystyle\int_0^1\frac{2t+i}{t^2+it^2+1}dt = \int_0^1\frac{2t^3+3t+i-it^2}{t^4+3t^2+1}dt =\int_0^1\frac{2t^3+3t}{t^4+3t^2+1}dt+i\int_0^1 \frac{1-t^2}{t^4+3t^2+1}dt$

I tried multiplying through by the conjugate but that didn't seem fruitful and left me with the above expression. Is there a better way to tackle this problem?

Typo I mixed up two parts

 

[Opalg]

$\displaystyle\int_0^1\frac{2t+i}{t^2+it^2+1}dt$

For a start, it looks as though there is a typo in the denominator. Shouldn't it be $\int_0^1\frac{2t+i}{t^2+it+1}dt$ ?

If so, the numerator is the derivative of the denominator, and the integral is
$\left[\ln(t^2+it+1)\right]_0^1 = \ln(2+i) = \sqrt5 + i\tan^{-1}\frac12$.

 

[Dustinsfl]

For a start, it looks as though there is a typo in the denominator. Shouldn't it be $\int_0^1\frac{2t+i}{t^2+it+1}dt$ ?

If so, the numerator is the derivative of the denominator, and the integral is
$\left[\ln(t^2+it+1)\right]_0^1 = \ln(2+i) = \sqrt5 + i\tan^{-1}\frac12$.

Shouldn't that be $\ln\sqrt{5} + i\tan^{-1}\frac{1}{2}$?

 

[Opalg]

Shouldn't that be $\ln\sqrt{5} + i\tan^{-1}\frac{1}{2}$?

Yes! (Doh)