Give the calculation in English plz

Let $u = 2-x^2$.
Then, differentiate $u$ with respect to $x$: $\frac{du}{dx} = -2x$.
Rearrange to find $dx$: $dx = \frac{du}{-2x}$.
Substitute $u$ and $dx$ into the integral:

The $x$ terms cancel out:

Integrate using the power rule $\int u^n du = \frac{u^{n+1}}{n+1} + C$:

Let $u = x^2$ and $dv = \cos 3x \, dx$.
Then $du = 2x \, dx$ and $v = \int \cos 3x \, dx = \frac{1}{3} \sin 3x$.

Let $u_1 = x$ and $dv_1 = \sin 3x \, dx$.
Then $du_1 = dx$ and $v_1 = \int \sin 3x \, dx = -\frac{1}{3} \cos 3x$.

The integral is $\boxed{\frac{1}{3} x^2 \sin 3x + \frac{2}{9} x \cos 3x - \frac{2}{27} \sin 3x + C}$.

The denominator is $x^2+2x-3 = (x+3)(x-1)$.
Set up the partial fractions:

Multiply both sides by $(x+3)(x-1)$:

To find $A$, set $x=-3$:

To find $B$, set $x=1$:

The integral is $\boxed{-10 \ln|x+3| + 4 \ln|x-1| + C}$.

Set the equations equal to each other:

Rearrange into a quadratic equation:

Factor the quadratic equation:

The points of intersection are $x = -1$ and $x = 2$. These will be the limits of integration.

Choose a test point within the interval, for example, $x=0$.
For the parabola: $y = (0)^2+4 = 4$.
For the line: $y = 0+6 = 6$.
Since $6 > 4$, the line $y=x+6$ is above the parabola $y=x^2+4$ in the interval $[-1, 2]$.

The area $A$ is given by the integral of the upper function minus the lower function:

Integrate term by term:

Apply the limits of integration:

Find a common denominator:

Simplify the fraction:

The area of the region is $\boxed{\frac{9}{2} \text{ square units}}$.