I have found the following problem:
By integrating both sides of an appropriate integral relationship, with suitable limits, show that: $$\int_0^1\frac{x^b-x^a}{\ln(x)}dx=\ln\left[\frac{b+1}{a+1}\right]$$ I started by saying: $$I_a=-\int_0^1x^adx=-\left[\frac{x^a}{a+1}\right]_0^1=-\frac{1}{a+1}$$ $$I_b=\int_0^1x^bdx=\left[\frac{x^b}{b+1}\right]_0^1=\frac{1}{b+1}$$ Which I believe is correct but I have not really dealt with partial differential equations so I have no idea how to proceed from here.
I believe from this we can say: $$I=-\ln|a+1|+C$$ $$I=\ln|b+1|+C$$ so can i just combine these
I'm not sure what you meant by $I_a$ or $I_b$, but I think you had something like this in mind: define \begin{equation} I(t) = \int_0^1 \dfrac{x^t}{\ln(x)} \, dx \end{equation} What you're being asked to compute is $I(b) -I(a)$. Let's first use the Leibniz integral rule to compute $I'(t)$: \begin{align} I'(t) &= \int_0^1 \dfrac{\partial}{\partial t} \left( \dfrac{x^t}{\ln(x)} \right) \, dx \\ &= \int_0^1 \dfrac{x^t \ln(x)}{\ln(x)} \, dx \\ &= \int_0^1 x^t \, dx \\ &= \dfrac{1}{t+1}. \end{align}
Therefore, \begin{align} I(b) - I(a) &= \int_a^b I'(t) \, dt \\ &= \int_a^b \dfrac{1}{t+1} \, dt \\ &= \ln \left( \dfrac{b+1}{a+1} \right), \end{align}