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November 01, 2025, 07:45:55 pm

Author Topic: in need of help :(  (Read 621 times)  Share 

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yang_dong

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in need of help :(
« on: August 22, 2014, 05:27:14 pm »
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how would i integrate:
(1/x)(sqrt((x^4+1)/x^4))
« Last Edit: August 22, 2014, 06:59:36 pm by yang_dong »

tlbdrummerkid

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Re: in need of help :(
« Reply #1 on: August 28, 2014, 09:43:37 am »
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Let x^2 =tanu
2xdx=(secu)^2du
2x^2dx/x= (secu)^2du
(1/x)dx= (secu)^2du/(2x^2)
(1/x)dx= (secu)^2du/(2tanu)
(1/x)dx= 1/2cotudu

∫(1/x)(sqrt((x^4+1)/x^4))dx
=∫((1/2)cotu)(sqrt(((tanu)^2+1)/(tanu)^2))du
=∫((1/2)cotu)(sqrt(((secu)^2)/(tanu)^2))du
=∫((1/2)cotu)(sqrt((cosecu)^2))du
=∫((1/2)cotu)(cosecu)du
=∫((1/2)secu)du
=∫((1/2)secu(secu+tanu)/ (secu+tanu))du
=∫((1/2)((secu)^2+secutanu)/ (secu+tanu))du

Let w=secu+tanu
dw=(secutanu+(secu)^2)du

∫((1/2)((secu)^2+secutanu)/ (secu+tanu))du
=∫((1/2)1/w)dw
=(1/2)ln(w)+c
=(1/2)ln(secu+tanu)

x^2=tanu
x^4=(tanu)^2
x^4+1=(secu)^2
secu=sqrt(x^4+1)

(1/2)ln(secu+tanu)
=(1/2)ln(sqrt(x^4+1)+x^2)+c
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