# Integration by parts vs integration by parts with substitution different answers

I'm trying to integrate

$$\int(5x^4+1)\ln(x^5+x)dx$$

When I use an online calculator (it first uses substitution and then integration by parts) I get the answer

$$(x^5+x)\ln(x^5+x)-x^5-x$$

but when I try a different approach (without substitution only with integration by parts) I get

$$(x^5+x)\ln(x^5+x)-x$$

(I change the variable of $dx$ to $x^5+x$)

• The first is correct modulo a constant of integration, the second is wrong. Without any more info, this is because you didn't do the second method correctly.
– Paul
Commented Jan 4, 2018 at 14:54
• Use Latex to type your question. Put all the math parts inside two dollar signs and use the command \int_{}^{} to display your intergral Commented Jan 4, 2018 at 14:54 • Is that integral supposed to be \int(5x^4+1)\ln(x^5+x)dx ? Commented Jan 4, 2018 at 15:00 • @PM2Ring yes that's it – john Commented Jan 4, 2018 at 15:02 ## 3 Answers It seem's that you forgot something since diferentiation of (x^5+x)\ln(x^5+x)-x gives: \begin{align*} D_x\left[(x^5+x)\ln(x^5+x)-x\right]&=(5x^4+1)\ln(x^5+x)+5x^4+1-1\\ &=(5x^4+1)\ln(x^5+x)+5x^4 \end{align*} And in the expression(x^5+x)\ln(x^5+x)-x the last $x$ must be $x^5+x$ because the change that you used in order to integrate, i.e. $x^5+x$ instead of $x$.

• oh it seems I didn't calculate the first derivate the right way... didn't see it was a function of a function...
– john
Commented Jan 4, 2018 at 15:19

Since your integral looks like $\int g’\ln(g)$ you just need to find a primitive of $\ln( x)$, namely $x\cdot\ln(x)-x$, and just plug in $x^5+x$ for $g$.

Integration with x only \begin{aligned} \int\left(5 x^4+1\right) \ln \left(x^5+x\right) d x&=\int \ln \left(x^5+x\right) d\left(x^5+x\right) \\ &=\left(x^5+x\right) \ln \left(x^5+x\right)-\int\left(x^5+x\right) \frac{5 x^4+1}{x^5+x} d x \\ &=\left(x^5+x\right) \ln \left(x^5+x\right)-\int\left(5 x^4+1\right) d x \\ &=\left(x^5+x\right) \ln \left(x^5+x\right)-x^5-x+C \end{aligned} Integration with a new variable u

Let $$u=x^5+x$$, then $$d u=\left(5 x^4+1\right) d x$$ \begin{aligned} \int\left(5 x^4+1\right) \ln \left(x^5+x\right) d x &=\int \ln u d u \\ &=u \ln u-\int u \cdot \frac{1}{u} d u \\ &=u \ln u-u+C \\ &=\left(x^5+x\right) \ln \left(x^5+x\right)-x^5-x+C \end{aligned}