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January 30, 2015

January 30, 2015

Posted by **Tracy** on Sunday, November 20, 2011 at 10:24am.

(e^3x)(cosh)(2x)(dx)

A.(1/2)(e^5x)+(1/2)(e^x)+C

B.(1/10)(e^5x)+(1/2)(e^x)+C

C.(1/4)(e^3x)+(1/2)(x)+C

D.(1/10)(e^5x)+(1/5)(x)+C

- CALCULUS -
**Damon**, Sunday, November 20, 2011 at 10:27amI did your last one by parts. Try the same trick again.

- CALCULUS -
**Damon**, Sunday, November 20, 2011 at 11:09am - CALCULUS -
**Damon**, Sunday, November 20, 2011 at 11:11amUse for function

e^(3x)*cosh(2x)

use for variable

x

- CALCULUS -
**Damon**, Sunday, November 20, 2011 at 1:44pm(e^3x)(cosh)(2x)(dx)

but cosh 2x = (1/2) (e^2x + e^-2x)

so

(1/2) [ e^5x + e^x ] dx

(1/10)e^5x + (1/2)e^x + c

so B

which we already knew from Wolfram

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