Question 534008
First problem:   {{{int(1/(3*x^10),dx) }}}
.
Bring the constant {{{1/3}}} outside the integration sign.
.
{{{(1/3)*int((1/x^10),dx) }}}
.
Recall that if a variable in the denominator has a positive exponent, it can be brought into the numerator with a negative exponent. So, the {{{x^10}}} in the denominator can be moved into the numerator as {{{x^(-10)}}} changing the problem to:
.
{{{(1/3)*int((x^(-10)),dx) }}}
.
Now integrate using the rule {{{int(u^n,dx) = u^(n+1)/(n+1)}}}
.
This rule results in the integration becoming:
.
{{{(1/3)*int((x^(-10)),dx)=(1/3)*(x^(-10+1)/(-10+1)) }}}
.
Simplify the answer by adding the -10+1 in two places to get the answer:
.
{{{(1/3)*(x^(-9)/(-9))}}}
.
Further simplify the answer by factoring out the -9 from the denominator and multiplying it times the 3 in the denominator of the multiplier {{{1/3}}} to make it:
.
{{{(-1/27)*(x^(-9))}}}
.
It would make things a little more conventional to move the variable x into the denominator by changing its exponent from -9 to +9 in doing so:
.
{{{((-1)/27)*(1/x^9)}}}
.
Finally, since this was an indefinite integration, don't forget to include the constant c in the answer ... thereby making the final answer:
.
{{{((-1)/27)*(1/x^9)+c}}}
.
You could also write this as:
.
{{{((-1)/(27x^9))+ c}}}
.
If you so choose to do so.
.
Second problem:
.
{{{int((3+4u)/u,du)}}}
.
Begin splitting this into two separate problems by dividing the u into the denominator into both terms of the numerator as follows:
.
{{{int(((3/u)+(4u/u)),du)}}}
.
Now multiply both of these two terms by du and you change the problem to:
.
{{{int((3/u),du)+int((4u/u),du)}}}
.
In the first integration move the constant 3 outside the integration sign and in the second integration move the constant 4 outside the integration sign. When you do those two moves you have:
.
{{{3*int((1/u),du)+4*int((u/u),du)}}}
.
You should recognize the first integration is of the standard form 
.
{{{int(1/x,dx) = ln(abs(x))}}}
.
Note that the x is an absolute value.
.
Completing the first integration and multiplying it by the constant 3 results in the problem being reduced to:
.
{{{3*ln(abs(u)) + 4*int((u/u),du)}}}
.
In the second integral notice that {{{u/u = 1}}} so you are left with the following:
.
{{{3*ln(abs(u)) + 4*int(1,du)}}}
.
and the integration of du results in just u. Since this is problem involves just indefinite integrals the constant c must be added and the result becomes:
.
{{{3*ln(abs(u)) + 4*u + c}}}
.
That's it. Hope that these two problems help you to understand better the process of integration or anti-derivation, whatever you have been taught to call it.
.