Question

the integral for 8x(x2+4) is 8(ln(x)/4−ln(x2+4)/8)+C
Simplify:
2ln(x)−ln(x2+4)+C
can anyone explain how we get the minus and where the 8 has gone?? :O

Answer 1

is it like this?\[\int\limits \frac{ 8 }{ x \left( x^2+4 \right) }dx\]

Answer 2

are you here?

Answer 3

yes it is

Answer 4

\[\frac{A}{x}+\frac{Bx+C}{x^2+4}=\frac{8}{x(x^2+4)} \\ A(x^2+4)+(Bx+C)x=8 \\ Ax^2+4A+Bx^2+Cx=8 \\ 4A=8 \\ C=0 \\ A+B=0 \\ \text{ find } A \text{ and } B \\ \text{ we already have } C=0\]

Answer 5

\[I=\int\limits \frac{ 8x }{ x^2\left( x^2+4 \right) }dx\]
put x^2=t
2x dx=dt
\[I=4\int\limits \frac{ dt }{ t \left( t+4 \right) }\]
\[=4\int\limits \left[ \frac{ 1 }{ t \left( 0+4 \right) }+\frac{ 1 }{ -4\left( t+4 \right) } \right]dt\]
\[=\int\limits \left( \frac{ 1 }{ t }-\frac{ 1 }{ t+4 } \right)dt\]
\[=\ln \left| t \right|-\ln \left| t+4 \right|+c\]
?

Answer 6

Thanx surjithayer! is that method has a name?

Answer 7

it is just practice and making fractions.

Answer 8

The method is known as "U substitution". http://tutorial.math.lamar.edu/Classes/CalcI/SubstitutionRuleDefinite.aspx