soal tes camaba , tolong bantu ya kk' mastah matematika

[tex]\displaystyle \int\limits^1_0\frac{x^7-1}{\ln x}\,dx=\int\limits^1_0\frac{x^7}{\ln x}-\frac{1}{\ln x}\,dx\\\int\limits^1_0\frac{x^7-1}{\ln x}\,dx=\int\limits^1_0\frac{x^8}{\ln x}\cdot\frac1x\,dx-\int\limits^1_0\frac{1}{\ln x}\,dx\\\int\limits^1_0\frac{x^7-1}{\ln x}\,dx=\int\limits^1_0\frac{x^8}{\ln x}\cdot\frac1x\,dx-\int\limits^1_0\frac{1}{\ln x}\,dx\\\\\text{misal :}\\\ln x=u\\\frac1x\,dx=du\\\\\int\frac{x^8}{\ln x}\cdot\frac{1}{x}\,dx=\int\frac{e^{8u}}{u}\,du\\\text{misal :}\\8u=v\\8\,du=dv[/tex]
[tex]\displaystyle \int\frac{x^8}{\ln x}\cdot\frac{1}{x}\,dx=\int\frac{e^{8u}}{u}\cdot\frac88\,du\\\int\frac{x^8}{\ln x}\cdot\frac{1}{x}\,dx=\int\frac{e^{8u}}{8u}\cdot8\,du\\\int\frac{x^8}{\ln x}\cdot\frac{1}{x}\,dx=\int\frac{e^{v}}{v}\,dv\\\int\frac{x^8}{\ln x}\cdot\frac{1}{x}\,dx=\text{E}_{\text{i}}(v)\\\int\frac{x^8}{\ln x}\cdot\frac{1}{x}\,dx=\text{E}_{\text{i}}(8u)\\\int\frac{x^8}{\ln x}\cdot\frac{1}{x}\,dx=\text{E}_{\text{i}}(8\ln x)\\\\\\\int\frac1{\ln x}\,dx=\text{li}(x)[/tex]


[tex]\displaystyle \int\limits^1_0\frac{x^7-1}{\ln x}\,dx=\left\text{E}_{\text{i}}(8\ln x)-\text{li}(x)\right|^1_0\\\boxed{\boxed{\int\limits^1_0\frac{x^7-1}{\ln x}\,dx=\ln8}}[/tex]