Thread: Substitution THEN Integration By Parts

1. Substitution THEN Integration By Parts

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.

2. 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.

3. 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 $dx = \frac{dy}{2x}$ and so the integral becomes:

$\int x^3 \ln y \, \frac{dy}{2x} = \frac{1}{2} \int x^2 \ln y \, dy = \frac{1}{2} \int (y + 1) \ln y \, dy$.

Now use parts.

4. Hello, SportfreundeKeaneKent!

$\int x^3\ln(x^2+1)\,dx$
By parts . . .

. . $\begin{array}{ccccccc}u &=& \ln(x^2+1) & \quad & dv &=& x^3\,dx \\ \\[-3mm] du &=&\dfrac{2x}{x^2+1}\,dx & \quad & v &=&\frac{1}{4}x^4 \end{array}$

We have: . $\frac{1}{4}x^4\ln(x^2+1) - \int\left(\frac{1}{4}x^4\right)\left(\frac{2x}{x^2 +1}\,dx\right)$

. . . . . . $= \;\frac{1}{4}x^4\ln(x^2+1) - \frac{1}{2}\int\frac{x^5}{x^2+1}\,dx$

. . . . . . $= \;\frac{1}{4}x^4\ln(x^2+1) - \frac{1}{2}\left(x^3 - x + \frac{x}{x^2+1}\right)\,dx$

Got it?

5. 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

6. 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