91Ó°ÊÓ

First, we need to simplify the integrand \(\frac{x}{x^2 + 2x + 2}\). To do so, complete the square of the denominator: \[x^2 + 2x + 2 = (x^2 + 2x + 1) + 1 = (x + 1)^2 + 1\] So, the integrand becomes: \[\frac{x}{(x + 1)^2 + 1}\]

Now we need to find the antiderivative of the simplified integrand. To do this, we will use substitution. Let \[u = x+1\] Hence, \[du = dx\] and \[\int \frac{x}{(x+1)^2 + 1} dx = \int \frac{u-1}{u^2 + 1} du\] Now, let's find the antiderivative of \(\frac{u-1}{u^2 + 1}\): \[\int \frac{u-1}{u^2 + 1} du = \int \left(\frac{1}{u^2 + 1}-\frac{1}{u^2 + 1}u\right) du\] The first term is the derivative of \(\arctan(u)\), whereas the second term can be computed using integration by parts. Let \(dv = \frac{u}{u^2+1}du\). Then, \(v = \int \frac{u}{u^2+1}du = \frac{1}{2}\ln(u^2+1)+C\). Let \(u = w\). Then, \(du = dw\). Thus: \[\int udv = uv - \int vdu = w\left(\frac{1}{2}\ln(w^2+1)+C\right) - \int \left(\frac{1}{2}\ln(w^2+1)+C\right) dw\] We know that: \[\int \frac{1}{w^2+1}dw = \arctan(w)\] and: \[\int \ln(w^2+1)dw = w\left(\ln(w^2+1)+C\right) - \int \left(\frac{1}{w^2+1}w(w^2+1)\right) dw\] So, our antiderivative is: \[F(x) = \arctan(u) - \frac{1}{2}\ln(u^2+1)(u-1) + C\] Now, substitute \(x+1\) back for \(u\): \[F(x) = \arctan(x+1) - \frac{1}{2}\ln((x+1)^2+1)(x+1-1) + C\]

Now we can apply the Fundamental Theorem of Calculus to evaluate the definite integral over the interval \([-1, 0]\): \[\int_{-1}^{0} \frac{x}{x^2+2x+2} dx = F(0) - F(-1) = \arctan(1) - \frac{1}{2}\ln(2) - (\arctan(0) - 0)\] \[= \frac{\pi}{4} - \frac{1}{2}\ln(2)\] Our final answer is: \[\int_{-1}^{0} \frac{x}{x^2+2x+2} dx = \boxed{\frac{\pi}{4} - \frac{1}{2}\ln(2)}\]