Antiderivative of $\log(x)$ without Parts

I understand how the antiderivative of $\log(x)$ can be obtained by Integration by Parts (i.e. product rule), but I was wondering how-if at all- it could be obtained only using sum/difference rule and substitution/chain rule.

Solution 1:

$$\int_1^t \ln(x)\,dx = \int_1^t\int_1^x\frac 1u\,du\,dx = \int_1^t\int_u^t\frac 1u\,dx\,du = \int_1^t\frac{t-u}{u}\,du = {\large[}t\ln(u)-u{\large]}_{1}^t$$

Solution 2:

There's a fun formula relating the integral of an invertible function and the integral of its inverse, namely $$ bf(b)-af(a)=\int_a^bf(t)dt+\int_{f(a)}^{f(b)}f^{-1}(t)dt $$ Which is easily seen from a picture. Then setting $b=x$, $0<x<a$ and $f(t)=\log t$ gives $$ x\log x-a\log a=\int_a^x\log tdt+\int_{\log a}^{\log x}e^tdt\\ \Rightarrow x\log x-a\log a-\int_{\log a}^{\log x}e^tdt=\int_a^x\log tdt\\ \Rightarrow x\log x-x+a-a\log a=\int_a^x\log tdt\\ \Rightarrow x\log x-x+c=\int_a^x\log tdt $$

Solution 3:

We have:

$$\int_0^{t} x^a dx = \frac{t^{a+1}}{a+1}$$

Differentiating w.r.t. $a$ gives:

$$\int_0^{t} x^a \log(x)dx = \frac{t^{a+1}\left[(a+1)\log(t)-1\right]}{(a+1)^2}$$

The result then follows if we put $a = 0$.