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