Question

integral of 3x^2 +3x +3/x^2 + 1

Answer 1

divide first

Answer 2

I did .. and got 3 + 3x/x^2 + 1

Answer 3

ok good

Answer 4

the anti derivative of \(3\) is \(3x\)

Answer 5

\[3\int\frac{xdx}{x^2+1}\] a mental u - sub gives the log

Answer 6

i.e. \(u=x^2+1, du =2xdx, \frac{du}{2}=xdx\) you get
\[\frac{3}{2}\int \frac{du}{u}=\frac{3}{2}\ln(u)=\frac{3}{2}\ln(x^2+1)\]

Answer 7

im lost .. after I divided I got \[\int\limits 3x + 3x \div x ^{2} + 1\]

Answer 8

ok maybe there was a mistake somewhere, lets see

Answer 9

\[\int\limits 3x ( 1 + \frac{ 1 }{ x ^{2}+1}\]

Answer 10

\[\frac{3x^2+3x+3}{x^2+1}=\frac{3(x^2+x+1)}{x^2+1}\]
\[=3\frac{x^2+x+1}{x^2+1}=3\left(\frac{x^2+1}{x^2+1}+\frac{x}{x^2+1}\right)\]

Answer 11

\[=3+\frac{3x}{x^2+1}\]

Answer 12

then
\[\int(3+\frac{3x}{x^2+1})dx=\int 3dx+\int \frac{3xdx}{x^2+1}\]
\[=3x+\frac{3}{2}\ln(x^2+1)\]

Answer 13

oh .. you pulled out the three in the beginning.

Answer 14

yes

Answer 15

but it makes no difference really
the trick is to divide it up in to two parts, one with the numeration of \(x^2+1\) or \(3x^2+3\) to give one term of 3, and the other of \(\frac{3x}{x^2+1}\)

Answer 16

but how did you get \[\frac{ 3 }{ 2 }\]

Answer 17

nevermind .. i see.