To evaluate the integral, we can split it into two integrals:

\(\displaystyle \int x^2 dx+ \frac{1}{2}\int \frac{dx}{x}\)

After integrating, we get

\(\displaystyle \frac{x^3}{3}+\frac{1}{2}\ln\left | x\right |+C\)

where a single constant of integration comes from the sum of the two integration constants from the two individual integrals, added together.

The rules used to integrate are

\(\displaystyle \int x^ndx=\frac{x^{n+1}}{n+1}+C\), \(\displaystyle \int \frac{dx}{x}=\ln\left | x\right | +C\)

The integral is equal to

\(\displaystyle \frac{1}{2\sqrt{3}}\arctan(\frac{3x}{2\sqrt{3}})+C\)

and was found using the following rule:

\(\displaystyle \int \frac{dx}{a^2+x^2}=\frac{1}{a}\arctan{\frac{x}{a}}+C\)

where \(\displaystyle a=\sqrt{12}=2\sqrt{3}\)

To integrate, we can split the integral into the sum of two separate integrals:

\(\displaystyle \int \frac{dx}{x^2}+\frac{1}{3}\int \frac{dx}{x}\)

Integrating, we get

\(\displaystyle -\frac{1}{x}+\frac{1}{3}\ln\left | x\right |+C\)

which was found using the following rules:

\(\displaystyle \int x^n dx=\frac{x^{n+1}}{n+1}+C\), \(\displaystyle \int \frac{dx}{x}=\ln\left | x\right |+C\)

Note that the constants of integration were combined to make a single integration constant in the final answer. 

(The first integral can be rewritten as  \(\displaystyle \int x^{-2}dx\) for clarity.)

Solving this integral depends on knowledge of exponent rules; mainly, that \(\displaystyle (a^b)^c = a^{b*c}\). Using this, we can simplify the given expression.

\(\displaystyle \int(e^2)^{\ln3x}dx = \int e^{2 * \ln 3x}dx = \int (e^{\ln 3x})^2dx =\int (3x)^2dx =\int 9x^2dx\)

From here, we just follow the power rule, raising the exponent and dividing by it.

\(\displaystyle 9\int x^2 dx = 3x^3 + C\)

Giving us our final answer.

To evaluate the integral, we use the following definition

\(\displaystyle \int x^n dx=\frac{x^{n+1}}{n+1}+C\)

\(\displaystyle \int (4x^4+2x^2+3x )dx=\frac{4x^5}{5}+\frac{2x^3}{3}+\frac{3x^2}{2}+C\)

To solve the problem, we apply the fact that anti-derivative of \(\displaystyle \sin(x)=-\cos(x)\) and that \(\displaystyle \int x^n dx=\frac{x^{n+1}}{n+1}+C\)

Taking the anti-derivative of each part independently, we get

\(\displaystyle \int \sin(x)dx+\int \frac{x}{4}dx\)

Finally, our answer is

 \(\displaystyle -\cos(x)+\frac{x^2}{8}+C\)

We can factor the equation inside the square root:

\(\displaystyle \\\int_0^2\sqrt{9x^2+12x+4}dx \\\\\int_0^2\sqrt{(3x+2)^2}dx \\\\\int_0^2 (3x+2)dx\)

From here, increase each term's exponent by one and divide the term by the new exponent.

\(\displaystyle \frac{3x^2}{2}+2x\big|_0^2\)

Now, substitute in the upper bound into the function and subtract the lower bound function value from it.

\(\displaystyle \\\frac{3(2)^2}{2}+2(2)-\left(\frac{3(0)^2}{2}+2(0)\right) \\\\\frac{3(4)}{2}+4 \\\\\frac{12}{2}+4 \\\\6+4 \\\\10\)

Therefore,

\(\displaystyle \int_{0}^{2}\sqrt{(3x+2)^2}dx=\int_{0}^{2}(3x+2)dx=10\)

To evaluate the integral, we use the definition 

\(\displaystyle \int x^n dx=\frac{x^{n+1}}{n+1}+C\)

\(\displaystyle \int (3x^3+2x-5 )dx=\frac{3}{4}x^4+x^2-5x+C\)

To evaluate the integral, we use the fact that the antiderivative of \(\displaystyle \sin{x}\) is \(\displaystyle -\cos{x}\) (because \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(-\cos{x})=\sin{x}\)), and the antiderivative of \(\displaystyle \sec^2{x}\) is \(\displaystyle \tan{x}\) (because \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(\tan{x})=\sec^2x\)). Using this information, we determine that the integral is

\(\displaystyle -\cos{x}+\tan{x}+C\)

Calculate the following integral.

\(\displaystyle \int4sin(t)-12cos(t)+3t^2dt\)

To do this problem, we need to recall that integrals are also called antiderivatives. This means that we can calculate integrals by reversing our integration rules.

Thus, we can have the following rules.

\(\displaystyle \int sin(t)dt = -cos(t)+c\)

\(\displaystyle \int cos(t)dt = sin(t)+c\)

\(\displaystyle \int t^ndt=\frac{t^{n+1}}{n+1}+c\)

Using these rules, we can find our answer:

\(\displaystyle \int4sin(t)-12cos(t)+3t^2dt\) 

Will become:

\(\displaystyle -4cos(t)-12sin(t)+\frac{3t^3}{3}+c=-4cos(t)-12sin(t)+t^3+c\)

And so our answer is:

\(\displaystyle -4cos(t)-12sin(t)+t^3+c\)