Keeping in mind that a and b are only constants, the integral is equal to

$\displaystyle \int abe^{b^2x}dx=\frac{abe^{b^2x}}{b^2}+C=\frac{ae^{b^2x}}{b}+C$

and was found using the following rule:

$\displaystyle \int e^{ax}dx=\frac{e^{ax}}{a}+C$

To evaluate the integral, we must make the following substitution:

$\displaystyle u=x^2y$, $\displaystyle du=2xydx$

The derivative was found using the following rule:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Rewrite the integral in terms of u and integrate:

$\displaystyle \frac{1}{2}\int \cos(u)du=\frac{1}{2}\sin(u)+C$

The integral was found using the following rule:

$\displaystyle \int \cos(x)dx=\sin(x)+C$

Finally, replace u with our original term:

$\displaystyle \frac{1}{2}\sin(x^2y)+C$.

Although this may seem really difficult, we only need to solve for $\displaystyle Q$. 

$\displaystyle \frac{d}{dx}V =\frac{d}{dt}Q*R$

$\displaystyle \frac{d}{dt}Q=\frac{\frac{\mathrm{d} }{\mathrm{d} x}V}{R}$

To solve for $\displaystyle Q$, integrate both sides:

$\displaystyle Q=\int \frac{\frac{d}{dx}V}{R} dt$

When integrating, remember to add one to the exponent and then put that result on the denominator: $\displaystyle \frac{2x^3}{3}-\frac{5x^2}{2}+x$. Now evaluate at 2, and then 0. Then subtract the two results. $\displaystyle (\frac{16}{3}-10+2)-0=-\frac{8}{3}$.

The first step here is to chop this up into three separate terms and then simplify since we have only one denominator: $\displaystyle \int 6x^2-\frac{1}{2}x+1dx$. Then, integrate each term, remembering to add one to the exponent and then put that result on the denominator: $\displaystyle 6(\frac{x^3}{3})-\frac{1}{2}(\frac{x^2}{2})+x$. Simplify to get your answer: $\displaystyle 2x^3-\frac{x^2}{4}+x+C$. Remember to add C because it is an indefinite integral.

First, chop this expression up into two terms: $\displaystyle \int \frac{9}{x}+1dx$. Then, integrate each term, remembering that when there is a single x on a denominator, the integral is $\displaystyle ln\left | x \right |$. Therefore, the integration is: $\displaystyle 9ln\left | x \right |+x+C$. Remember to add C because it is an indefinite integral.

First, integrate each term separately. Remember, when integrating, raise the exponent by one and then also put that result on the denominator: $\displaystyle 2(\frac{x^4}{4})-3(\frac{x^3}{3})+\frac{x^2}{2}$. Then evaluate at 3 and then 0. Subtract two results to get: $\displaystyle (\frac{81}{2}-27+\frac{9}{2})-0=18$.

To integrate this expression, I would first rewrite it to be $\displaystyle \int x^{\frac{1}{2}}dx$. Then, add one to the exponent and then put that result on the denominator: $\displaystyle \frac{x^{\frac{3}{2}}}{\frac{3}{2}}$. Then, simplify to get your answer of $\displaystyle \frac{2}{3}x^{\frac{3}{2}}+C$. Remember to add C because it is an indefinite integral.

Before we solve the problem by computation, let's look at the answer choices and see if we can eliminate any answer choices. We know that an inverse function must exist because f(x) is one-to-one, so we can eliminate the answer choice "undefined." Next, we know that the inverse function has to be in terms of y, so we can eliminate the two answer choices with an "x." 

Now we can look at the two remaining answer choices.  Let y=f(x) and solve for x to find the inverse.

So f(x)=y=3x+1. Solve for x.

x=(y-1)/3

Therefore our answer is (y-1)/3.

The polar equation formula here is x²+y²=r².

This is the formula for a cirlce, so we can eliminate the two "line" answers.

Plugging in r=-6 into the equation gives x²+y²=36, which describes the graph of a circle with radius 6.