- Integral Fungsi Trigonometri - 3 soal ya sahabat Brainly :) Jangan copas dan spam ya:)

[tex]\displaystyle 1.\\\text{misal :}\\\ln x=u\\\frac1x\,dx=du\\\\\int\frac{dx}{x(1+\ln^2x)}=\int\frac{1}{1+u^2}\,du\\\int\frac{dx}{x(1+\ln^2x)}=\arctan u+C\\\boxed{\boxed{\int\frac{dx}{x(1+\ln^2x)}=\arctan(\ln x)+C}}[/tex]


[tex]\displaystyle 2.\\\text{misal :}\\e^x=u\\e^x\,dx=du\\\\\int e^x\tan^{-1}e^x\,dx=\int \tan^{-1}u\,du\\\int e^x\tan^{-1}e^x\,dx=\int \arctan u\,du\\\int e^x\tan^{-1}e^x\,dx=\int \arctan u\cdot1\,du\\\int e^x\tan^{-1}e^x\,dx=\int \arctan u\,du\\\int e^x\tan^{-1}e^x\,dx=u\cdot\arctan u-\int\frac{u}{1+u^2}\,du\\\int e^x\tan^{-1}e^x\,dx=u\arctan u-\frac12\ln|1+u^2|+C\\\boxed{\boxed{\int e^x\tan^{-1}e^x\,dx=e^x\arctan e^x-\frac12\ln|1+e^{2x}|+C}}[/tex]

[tex]\displaystyle 3.\\\int x\tan^{-1}x\,dx=\int \tan^{-1}x\cdot x\,dx\\\int x\tan^{-1}x\,dx=\frac12x^2\cdot\tan^{-1}x-\int\frac12x^2\cdot\frac{1}{1+x^2}\,dx\\\int x\tan^{-1}x\,dx=\frac12x^2\cdot\tan^{-1}x-\frac12\int\frac{x^2+1-1}{1+x^2}\,dx\\\int x\tan^{-1}x\,dx=\frac12x^2\cdot\tan^{-1}x-\frac12\int1-\frac{1}{1+x^2}\,dx\\\int x\tan^{-1}x\,dx=\frac12x^2\cdot\tan^{-1}x-\frac12(x-\arctan x)+C\\\boxed{\boxed{\int x\tan^{-1}x\,dx=\frac12x^2\tan^{-1}x+\frac12\arctan x-\frac12x+C}}[/tex]