Nice integral

Consider. $I =\displaystyle \int_0^1 \dfrac{\ln(1+x^p)dx}{x}$ as, $\ln(1+x) = \displaystyle \sum_{n=1} ^ {\infty} \dfrac{ x^n(-1)^{n+1}}{n}$ plugging $x \rightarrow x^p$ $\ln(1+x^p) = \displaystyle \sum_{n=1} ^ {\infty} \dfrac{ x^np(-1)^{n+1}}{n}$ Note that as the integral has the limits from $0$ to $1$ so the above series is valid.
Hence $I =\displaystyle \int_0 ^1 \sum_{n=1} ^ {\infty} \dfrac{ x^np (-1)^{n+1} dx}{nx} =\displaystyle \sum_{n=1} ^ {\infty} \dfrac{(-1)^{n+1}}{n} \int_{0} ^ {1} x^{np-1} dx = \displaystyle \sum_{n=1} ^ {\infty} \dfrac{(-1)^{n+1}}{pn^2 }$
The second step is verified by Fubini-Tonelli's theorem
By the fact that $\displaystyle \sum_{n=1} ^ {\infty} \dfrac{(-1)^{n+1}}{n^2} = \dfrac{\pi^2}{12}$ Hence, $I= \dfrac{\pi^2}{12p}$ For our case , plug $p=4$ , so $I= \dfrac{\pi^2}{48}$ Clearly, $\phi(48-2)=\phi(46) = \boxed{\boxed{22}}$ $\large{ \text{Q.E.D} }$