Hi everyone, Can you tell me how to integrate the following equation? [tex]\int\frac{1}{x^2 + 1} \ dx[/tex] I've tried the substitution method, u = x^2 + 1, du/dx = 2x. But the x variable is still exist. Also, the trigonometry substitution method, but the denominator is not in [tex]\sqrt{x^2 + 1}[/tex] form. Thanks in advance Huygen
Wait -- are you saying you tried it and failed? (If so, would you show your work, please?) Or are you saying you didn't actually try it at all?
OK, as you asked for it: FIRST u = x^2 + 1, du/dx = 2x, du/2x = dx [tex]\int\frac{1}{x^2+1} \ dx = \int\frac{1}{u} \ \frac{du}{2x}[/tex] I can not do further because the x variable is still exist. SECOND [tex]x = tan \ \theta[/tex] [tex]\frac{dx}{d\theta}=sec^2\theta[/tex] [tex]dx=sec^2\theta \ d\theta[/tex] [tex]x^2+1=(tan \ \theta)^2+1=tan^2\theta+1=sec \ \theta[/tex] Then [tex]\int\frac{1}{x^2+1} \ dx=\int\frac{1}{sec \ \theta} \ (sec^2\theta \ d\theta) = \int sec \ \theta \ d\theta= ln(tan \ \theta+sec \ \theta) \ + \ C=ln[x+(x^2+1)]+C=ln(x^2+x+1)+C[/tex]
If you applied the correct identity it might help . . . [tex]tan^2(x) + 1 = sec^2(x)[/tex] Your integral equation then becomes, [tex]\int\frac{1}{x^2 + 1} \; dx = \int \, d\theta[/tex]
If you don't recognize the anti-derivative at first sight, the substitution allows you to evaluate the integral without too much difficulty.
hi, This is an old thread.. So if u'd like me to post it elsewhere, do let me know.. I need to calculate the theta using inverse-tan. But since micro-controllers do not provide much computational freedom, I was looking to solve it as the integral of 1/(1+x^2). Other than the fact that, integral of 1/(1+x^2) is arctan(x). But since i'd like to know arctan(x), could someone please help me to find the intergral in terms of x (non-trigonometric).
hi vish_al210! (try using the X^{2} icon just above the Reply box ) you could try expanding 1/(1+x^{2}) as 1 - x^{2} - … , and then integrating
Re: How to Integrate[1/(x^2 + 1)] dx? there is a direct formula for these kind of questins i.e integrate[1/(a^2+x^2)]dx = 1/a[arctan(x/a)] so u can assume a=1.....so ur ans vil b (arctan x)......jst dis..
welcome to pf! hi kanika2217! welcome to pf! yes we know, but we're all trying to do it the way the ancient greeks would have done! (btw, please don't use txt spelling on this forum … it's against the forum rules )
hi! v olso have a drct frmila for these kind of ques...i.e.integration[1/(a^2 + x^2)]dx = (1/a)(arctan (x/a)...so you can assume a=1 here and hence the ans wud b 'arctan x'....
Welcome to PF! Hi M1991! Welcome to PF! (try using the X^{2} button just above the Reply box ) no, that's ln(x^{2} + 1) - ln(2) - ln(x) … its derivative is 2x/(x^{2} + 1) - 1/x