- #1
frozen7
- 163
- 0
[tex] \int \sec hx[/tex]
I solve it in this way:
[tex] \int \arccos hx[/tex]
[tex] \int \ln (x^2 + \sqrt{x^2 -1}) dx [/tex]
Then, I substitute [tex] u = \ln (x + \sqrt{x^2 + 1})[/tex]
then I get
[tex] x\ln(x + \sqrt{x^2 + 1}) - \int x/\sqrt{x^2 + 1}dx[/tex]
and then I substitute v = x^2 + 1
[tex] x\ln(x + \sqrt{x^2 + 1}) -1/2 \int v^(1/2) dv [/tex]
and finally I get the answer,
[tex] x\ln(x + \sqrt{x^2 + 1}) -\sqrt{x^2 + 1} + C [/tex]
Am I right?
I solve it in this way:
[tex] \int \arccos hx[/tex]
[tex] \int \ln (x^2 + \sqrt{x^2 -1}) dx [/tex]
Then, I substitute [tex] u = \ln (x + \sqrt{x^2 + 1})[/tex]
then I get
[tex] x\ln(x + \sqrt{x^2 + 1}) - \int x/\sqrt{x^2 + 1}dx[/tex]
and then I substitute v = x^2 + 1
[tex] x\ln(x + \sqrt{x^2 + 1}) -1/2 \int v^(1/2) dv [/tex]
and finally I get the answer,
[tex] x\ln(x + \sqrt{x^2 + 1}) -\sqrt{x^2 + 1} + C [/tex]
Am I right?
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