3. $int_{0}^{1}\frac{1 + x}{1 + x^{2}}dx$ 4. $int_{-4}^{-2}\frac{dx}{-t^{2}-dt - 5}$

Explanation:

Step1: Split the first integral

Split $\int_{0}^{1}\frac{1 + x}{1+x^{2}}dx$ into $\int_{0}^{1}\frac{1}{1 + x^{2}}dx+\int_{0}^{1}\frac{x}{1+x^{2}}dx$.

Step2: Evaluate $\int_{0}^{1}\frac{1}{1 + x^{2}}dx$

We know that $\int\frac{1}{1 + x^{2}}dx=\arctan(x)+C$. So $\int_{0}^{1}\frac{1}{1 + x^{2}}dx=\arctan(1)-\arctan(0)=\frac{\pi}{4}-0=\frac{\pi}{4}$.

Step3: Evaluate $\int_{0}^{1}\frac{x}{1+x^{2}}dx$

Let $u = 1 + x^{2}$, then $du=2xdx$. When $x = 0$, $u = 1$; when $x = 1$, $u=2$. So $\int_{0}^{1}\frac{x}{1+x^{2}}dx=\frac{1}{2}\int_{1}^{2}\frac{du}{u}=\frac{1}{2}[\ln(u)]_{1}^{2}=\frac{1}{2}(\ln(2)-\ln(1))=\frac{1}{2}\ln(2)$.

Step4: Combine the results of the first - integral

$\int_{0}^{1}\frac{1 + x}{1+x^{2}}dx=\frac{\pi}{4}+\frac{1}{2}\ln(2)$.

Step5: Rewrite the second integral

The second integral $\int_{-4}^{-2}\frac{dt}{-t^{2}-t - 5}=-\int_{-4}^{-2}\frac{dt}{t^{2}+t + 5}$. Complete the square for the denominator: $t^{2}+t + 5=(t+\frac{1}{2})^{2}+5-\frac{1}{4}=(t + \frac{1}{2})^{2}+\frac{19}{4}$.

Step6: Use the integral formula $\int\frac{1}{a^{2}+x^{2}}dx=\frac{1}{a}\arctan(\frac{x}{a})+C$

Let $x=t+\frac{1}{2}$ and $a=\frac{\sqrt{19}}{2}$, then $-\int_{-4}^{-2}\frac{dt}{t^{2}+t + 5}=-\int_{-4+\frac{1}{2}}^{-2+\frac{1}{2}}\frac{dt}{(t+\frac{1}{2})^{2}+\frac{19}{4}}=-\frac{2}{\sqrt{19}}\left[\arctan\left(\frac{2t + 1}{\sqrt{19}}\right)\right]_{-\frac{7}{2}}^{-\frac{3}{2}}$. $=-\frac{2}{\sqrt{19}}\left(\arctan\left(\frac{-3 + 1}{\sqrt{19}}\right)-\arctan\left(\frac{-7+1}{\sqrt{19}}\right)\right)=-\frac{2}{\sqrt{19}}\left(\arctan\left(-\frac{2}{\sqrt{19}}\right)-\arctan\left(-\frac{6}{\sqrt{19}}\right)\right)$.

Answer:

The value of $\int_{0}^{1}\frac{1 + x}{1+x^{2}}dx$ is $\frac{\pi}{4}+\frac{1}{2}\ln(2)$ and the value of $\int_{-4}^{-2}\frac{dt}{-t^{2}-t - 5}=-\frac{2}{\sqrt{19}}\left(\arctan\left(-\frac{2}{\sqrt{19}}\right)-\arctan\left(-\frac{6}{\sqrt{19}}\right)\right)$