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.