First, chop up the fraction into two separate terms:

$\displaystyle \int \frac{5}{x}+xdx$

Now, integrate:

$\displaystyle 5ln\left | x \right |+\frac{x^2}{2}$

Evaluate at 4 and then at 1. Subtract the results:

$\displaystyle (5ln4+8)-(5ln1+\frac{1}{2})$

Round to four places:
$\displaystyle 14.9315-.5=14.4315$

First off, chop up the fraction into three separate terms:

$\displaystyle \int x^2-\frac{1}{5}x+2dx$

Integrate. Remember to raise the exponent by 1 and then also put that result on the denominator:

$\displaystyle \frac{x^3}{3}-\frac{1}{5}(\frac{x^2}{2})+2x$

Simplify and add C because it is an indefinite integral:

$\displaystyle \frac{x^3}{3}-\frac{x^2}{10}+2x+C$

First off, chop up the fraction into two separate terms:

$\displaystyle \int \frac{7}{x}+1dx$

Now integrate:

$\displaystyle 7lnx+x$

Evaluate at 5 and then 1. Subtract the results.

$\displaystyle (7ln5+5)-(7ln1+1)$

Round to four places:

$\displaystyle 15.2661$

Step 1: Rewrite the denominator as x to a certain power. A square root means that the exponent will have a value of $\displaystyle \frac {1}{2}$.

We get $\displaystyle x^{\frac {1}{2}}$.

Step 2: Divide the numerator and denominator of the function. When we divide terms with exponents, we must make sure that the bases are the same. Both bases are $\displaystyle x$, so let's continue. When you divide exponents, you are subtracting them. You subtract the bottom exponent FROM the top exponent. 
We get $\displaystyle 2-\frac {1}{2}$. We will convert the $\displaystyle 2$ into a fraction with denominator $\displaystyle 2$. By default, the denominator is $\displaystyle 1$. We will multiply the numerator and denominator by 2, so the new fraction is $\displaystyle \frac {4}{2}$.

Step 3: Subtract the converted fraction (exponent of top) and the exponent of the bottom.

$\displaystyle \frac {4}{2}-\frac {1}{2}=\frac {4-1}{2}=\frac {3}{2}$.

The exponent of the $\displaystyle x$ term is $\displaystyle \frac {3}{2}$.

Step 4: Integrate...
When we Integrate, we add $\displaystyle 1$ to the exponent. We then divide the new function with a new exponent by that new exponent.

So, we get: $\displaystyle \frac {x^{\frac {3}{2}+1}}{\frac {3}{2}+1}$.

Evaluate $\displaystyle \frac {3}{2}+1$ and then rewrite..
$\displaystyle \frac {3}{2}+1=\frac {3}{2}+\frac {2}{2}=\frac {3+2}{2}=\frac {5}{2}$

We get: $\displaystyle \frac {x^{\frac {5}{2}}}{\frac {5}{2}}$

Step 5: Evaluate the upper and the lower limit:

If $\displaystyle x=4:$

$\displaystyle \frac {4^{\frac {5}{2}}}{\frac {5}{2}}=\frac {\sqrt{4^5}}{\frac {5}{2}}=\frac {\sqrt{1024}}{\frac {5}{2}}=\frac {32}{\frac {5}{2}}$
Flip the denominator and multiply:

$\displaystyle 32 \cdot \frac {2}{5}=\frac {64}{5}$

If $\displaystyle x=1$
$\displaystyle \frac{1^\frac{5}{2}}{\frac{5}{2}}=\frac{1}{\frac{5}{2}}=\frac{1}{1}\times\frac{2}{5}=\frac{2}{5}$

Step 6:

Subtract lower limit from the upper limit:

$\displaystyle \frac{64}{5}-\frac{2}{5}=\frac{62}{5}$

This integral cannot be evaluated using the Fundamental Theorem of Calculus, since an antiderivative does not exist for $\displaystyle \sqrt{9-x^2}$. Instead, we need to find the area under the curve $\displaystyle y=\sqrt{9-x^2}$ bounded by the $\displaystyle x-$axis.

$\displaystyle y=\sqrt{9-x^2}$ from $\displaystyle -3< x< 3$, is the graph of the upper half of the circle $\displaystyle y^2+x^2=3^2$. The area of this upper half is

$\displaystyle \frac{\pi r^2}{2} = \frac{\pi(3)^2}{2} = \frac{9\pi}{2}$.

Hence $\displaystyle \int_{-3}^3 \sqrt{9-x^2}dx=\frac{9\pi}{2}$. 

First, chop up the fraction into two separate fractions:

$\displaystyle \int \frac{2}{x}+1dx$

Now, integrate:

$\displaystyle 2ln\left | x \right |+x$

Next, evaluate at 2 and then at 0. Subtract the results:

$\displaystyle 2ln(2)+2-(2ln(0)+0)=3.3863$.

First, integrate. Remember to raise the exponent by 1 and also put that result on the denominator:

$\displaystyle \frac{x^3}{3}-\frac{4x^2}{2}$

Simplify to get:

$\displaystyle \frac{x^3}{3}-2x^2$

Now, evaluate at 3 and then at 1. Subtract the results:
$\displaystyle (9+18)-(\frac{1}{3}+2)=27-(2\frac{1}{3})=\frac{74}{3}$

First, integrate. Remember to raise the exponent by 1 and also put that result on the denominator:

$\displaystyle \frac{x^{\frac{9}{5}}}{\frac{9}{5}}$

Simplify to get:

$\displaystyle \frac{5}{9}x^{\frac{9}{5}}$

Now, evaluate at 3 and then 0. Subtract the results:

$\displaystyle (\frac{5}{9}(3)^{\frac{9}{5}})-(\frac{5}{9}(0)^{\frac{9}{5}})=4.0137$

First, integrate. Remember to raise the exponent by 1 and also put that result on the denominator:

$\displaystyle \frac{3x^3}{3}+\frac{4x^2}{2}-x=x^3+2x^2-x$

Now, evaluate at 2 and then 1:

$\displaystyle (8+8-2)-(1+2-1)=14-2=12$.

Recall that when integrating, you must raise the exponent by 1 and also put that result on the denominator:

$\displaystyle \frac{x^{\frac{8}{7}}}{\frac{8}{7}}$

Simplify to get:
$\displaystyle \frac{7}{8}x^{\frac{8}{7}}$

Now, evaluate at 1 and then at 0:

$\displaystyle (\frac{7}{8}1^{\frac{8}{7}})-(\frac{7}{8}0^{\frac{8}{7}})=\frac{7}{8}$