Question

integrate by parts :
x tan ^-1 x
@SolomonZelman

Answer 1

i ended up here : x^2 /2 tan^-1 x - § 1 / root (x^2 + 1) * x^2 /2 .dx
and i lost it..

Answer 2

this is very nasty for me to explain, because I am myself a student...
§ x tan \(\scriptsize\color{black}{ ^{-1}}\)x dx

I don't see any other method to use other than your's :)

tan \(\scriptsize\color{black}{ ^{-1}}\)x=f
df = 1/(x²+1) dx
dg=xdx
g=x²/2

and this all is in `§f dg= fg - §g df `

Answer 3

This is how I would start at least.

Answer 4

m sorry but here we need to solve it "by parts" and not by substitution!

Answer 5

No, I am integrating ' x tan \(\scriptsize\color{black}{ ^{-1}}\)x ' like this, and it is by parts.

Answer 6

§f1 . f2 .dx= f1 §f2 dx - § [df1/dx §f2 dx ] dx
i meant solve it by this!

Answer 7

I am very confusing often times, but when I am continuing I am getting,

½ x² tan \(\scriptsize\color{black}{ ^{-1}}\)x - § x² / [ 2(x² +1) ] dx
½ x² tan \(\scriptsize\color{black}{ ^{-1}}\)x - ½§ x² / [ (x² +1) ] dx

then, long division for the second integral,

½ x² tan \(\scriptsize\color{black}{ ^{-1}}\)x - ½§ 1- [1/(x² +1) ] dx

Answer 8

½ x² tan \(\scriptsize\color{black}{ ^{-1}}\)x - ½§ 1/(x² +1) dx - ½ § 1 dx

Answer 9

½ x² tan\(\scriptsize\color{black}{ ^{-1}}\)(x) + ½ tan\(\scriptsize\color{black}{ ^{-1}}\)(x) -x/2 +C

Answer 10

then factor out of 1/2

Answer 11

is the differentiation of tan^-1 x = x^2 +1 or root ( x^2 +1 ) ?

Answer 12

none

Answer 13

1/ (x² +1)

Answer 14

yeah, .. I am not a very good helper.... igtg

Answer 15

ohk.. thankyou soo much this was my mistake..

Answer 16

okay thanks.. i got the answer (: and ur a great great helper :D

Answer 17

yw... btw, the highest math course I took was trigonometry, will leave it with this... bye and yw.

Answer 18

bbye (: @SolomonZelman