Evaluate integral ( x x)/(a 2 x + b 2 x) dx and Compute the value of integral (1+x2)/(1+x4) dx in terms of x.

*a) Evaluate $\int \frac{\sin x \cos x}{a \cos^2 x + b \sin^2 x} dx$.

Step 1: Use substitution. Let $u = a \cos^2 x + b \sin^2 x$. Now, find the differential $du$: $\frac{du}{dx} = \frac{d}{dx}(a \cos^2 x + b \sin^2 x)$ Using the chain rule, $\frac{d}{dx}(\cos^2 x) = 2 \cos x (-\sin x) = -2 \sin x \cos x$ and $\frac{d}{dx}(\sin^2 x) = 2 \sin x (\cos x)$. $\frac{du}{dx} = a(-2 \sin x \cos x) + b(2 \sin x \cos x)$ $\frac{du}{dx} = 2 \sin x \cos x (b - a)$ So, $du = 2(b-a) \sin x \cos x \, dx$.

Step 2: Express $\sin x \cos x \, dx$ in terms of $du$. Assuming $b \neq a$: $\sin x \cos x \, dx = \frac{1}{2(b-a)} du$

Step 3: Substitute $u$ and $du$ into the integral. $\int \frac{1}{u} \cdot \frac{1}{2(b-a)} du$ $= \frac{1}{2(b-a)} \int \frac{1}{u} du$

Step 4: Integrate with respect to $u$. $= \frac{1}{2(b-a)} \ln|u| + C$

Step 5: Substitute back $u = a \cos^2 x + b \sin^2 x$. $= \frac{1}{2(b-a)} \ln|a \cos^2 x + b \sin^2 x| + C$ The value of the integral is $\boxed{\frac{1}{2(b-a)} \ln|a \cos^2 x + b \sin^2 x| + C}$. (Note: If $b=a$, the integral becomes $\int \frac{\sin x \cos x}{a} dx = \frac{1}{a} \int \sin x \cos x dx$. Let $v = \sin x$, $dv = \cos x dx$. Then $\frac{1}{a} \int v dv = \frac{1}{a} \frac{v^2}{2} + C = \frac{\sin^2 x}{2a} + C$.)

*b) Compute the value of $\int \frac{1+x^2}{1+x^4} dx$ in terms of $x$.

Step 1: Divide the numerator and denominator by $x^2$. $\int \frac{\frac{1}{x^2}(1+x^2)}{\frac{1}{x^2}(1+x^4)} dx = \int \frac{\frac{1}{x^2}+1}{\frac{1}{x^2}+x^2} dx$ $= \int \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}} dx$

Step 2: Make a substitution. Let $u = x - \frac{1}{x}$. Then, find the differential $du$: $du = \frac{d}{dx}\left(x - \frac{1}{x}\right) dx = \left(1 - (-\frac{1}{x^2})\right) dx = \left(1 + \frac{1}{x^2}\right) dx$ Now, express the denominator $x^2 + \frac{1}{x^2}$ in terms of $u$: $u^2 = \left(x - \frac{1}{x}\right)^2 = x^2 - 2(x)\left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^2$ $u^2 = x^2 - 2 + \frac{1}{x^2}$ So, $x^2 + \frac{1}{x^2} = u^2 + 2$.

Step 3: Substitute $u$ and $du$ into the integral. $\int \frac{du}{u^2 + 2}$

Step 4: Integrate with respect to $u$. This is a standard integral of the form $\int \frac{1}{y^2+a^2} dy = \frac{1}{a} \arctan\left(\frac{y}{a}\right) + C$. Here, $y=u$ and $a^2=2$, so $a=\sqrt{2}$. $= \frac{1}{\sqrt{2}} \arctan\left(\frac{u}{\sqrt{2}}\right) + C$

Step 5: Substitute back $u = x - \frac{1}{x}$. $= \frac{1}{\sqrt{2}} \arctan\left(\frac{x - \frac{1}{x}}{\sqrt{2}}\right) + C$ Simplify the argument of the arctan function: $= \frac{1}{\sqrt{2}} \arctan\left(\frac{\frac{x^2-1}{x}}{\sqrt{2}}\right) + C$ $= \frac{1}{\sqrt{2}} \arctan\left(\frac{x^2-1}{\sqrt{2}x}\right) + C$ The value of the integral is $\boxed{\frac{1}{\sqrt{2}} \arctan\left(\frac{x^2-1}{\sqrt{2}x}\right) + C}$.