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.
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.
Hello, SportfreundeKeaneKent!
By parts . . .$\displaystyle \int x^3\ln(x^2+1)\,dx$
. . $\displaystyle \begin{array}{ccccccc}u &=& \ln(x^2+1) & \quad & dv &=& x^3\,dx \\ \\[-3mm] du &=&\dfrac{2x}{x^2+1}\,dx & \quad & v &=&\frac{1}{4}x^4 \end{array}$
We have: .$\displaystyle \frac{1}{4}x^4\ln(x^2+1) - \int\left(\frac{1}{4}x^4\right)\left(\frac{2x}{x^2 +1}\,dx\right) $
. . . . . .$\displaystyle = \;\frac{1}{4}x^4\ln(x^2+1) - \frac{1}{2}\int\frac{x^5}{x^2+1}\,dx $
. . . . . .$\displaystyle = \;\frac{1}{4}x^4\ln(x^2+1) - \frac{1}{2}\left(x^3 - x + \frac{x}{x^2+1}\right)\,dx $
Got it?