Question

definite integral 1+2x/1+x^2
upper limit 1
lower limit 0

Answer 1

Should I integrate it by parts or how should I start?

Answer 2

split it into two terms

Answer 3

\[\int\limits_{1}^{0} \frac{ 1+2x }{ 1+x^2 }\]
is the integral to simplify

Answer 4

@ganeshie8 can you help me please

Answer 5

okey so u = 1+x^2 and dv = 1+2x?

Answer 6

\[\large \int\limits_{0}^{1} \frac{ 1+2x }{ 1+x^2 } dx= \int\limits_{0}^{1} \frac{ 1}{ 1+x^2 }dx + \int\limits_{0}^{1} \frac{ 2x }{ 1+x^2 } dx \]

Answer 7

first integral evaluates to arctan(x),
use substitution for the second integral

Answer 8

okey hold on let I'll try.

Answer 9

\[u = 1+x^2 \]
du/dx = 2x
du /2 = x dx
so is it
\[\int\limits_{0}^{1} \frac{ du }{ u }\] ??

Answer 10

looks good^^

Answer 11

Yes, okey hold on:)

Answer 12

\[\left[ \ln x^2 +1 \right]\]
upper limit 1
lower limit 0
is it correct then just add the two equations together?

Answer 13

yes evaluate the bounds

Answer 14

yup I got the right answer.

Answer 15

good :)

Answer 16

thanx closing the thread. :)