First I have to use substitution on this, then integration by parts. ∫ x^3 ln(x^2 + 1) dx My guess is that y = x^2 + 1 Then dy = 2x dx but then I'm stuck since 2x isn't a 3 degree exponent like in the original integral.
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Originally Posted by SportfreundeKeaneKent First I have to use substitution on this, then integration by parts. ∫ x^3 ln(x^2 + 1) dx My guess is that y = x^2 + 1 Then dy = 2x dx but then I'm stuck since 2x isn't a 3 degree exponent like in the original integral. No substitution here, just integration by parts.
Originally Posted by SportfreundeKeaneKent First I have to use substitution on this, then integration by parts. ∫ x^3 ln(x^2 + 1) dx My guess is that y = x^2 + 1 Then dy = 2x dx but then I'm stuck since 2x isn't a 3 degree exponent like in the original integral. Therefore and so the integral becomes: . Now use parts.
Hello, SportfreundeKeaneKent! By parts . . . . . We have: . . . . . . . . . . . . . Got it?
That way is simpler Soroban but on a test when it says use sub first then I won't be allowed to just use integration by parts. Btw Mr_Fantastic, wouldn't the last line on your answer be 1/2∫(y-1)lny dy. y=x^2 + 1 so x^2 = y-1
Originally Posted by SportfreundeKeaneKent That way is simpler Soroban but on a test when it says use sub first then I won't be allowed to just use integration by parts. Btw Mr_Fantastic, wouldn't the last line on your answer be 1/2∫(y-1)lny dy. y=x^2 + 1 so x^2 = y-1 Correct. Thankyou for realising it was a mistake instead of panicking and saying that you're all confused now and just don't know what to do
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