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Question Number 46129 by Necxx last updated on 21/Oct/18

Commented by Meritguide1234 last updated on 21/Oct/18

$$putx=tan\theta$$$${\int }_0^{\pi /4}log(1+tan\theta )d\theta$$$$usef(a-x)\to f\left(x\right)$$$$I=\frac{\pi }{8}log2$$

Commented by maxmathsup by imad last updated on 21/Oct/18

$$changementx=tan\theta giveI={\int }_0^{\frac{\pi }{4}}\frac{ln(1+tan\theta )}{1+{tan}^2\theta }(1+{tan}^2\theta )d\theta$$$$={\int }_0^{\frac{\pi }{4}}ln\left(\frac{cos\theta +sin\theta }{cos\theta }\right)d\theta ={\int }_0^{\frac{\pi }{4}}ln\left(\sqrt{2}cos(\theta -\frac{\pi }{4})\right)d\theta -{\int }_0^{\frac{\pi }{4}}ln\left(cos\theta \right)d\theta$$$$=\frac{\pi }{8}ln\left(2\right)+{\int }_0^{\frac{\pi }{4}}cos(\theta -\frac{\pi }{4})d\theta -{\int }_0^{\frac{\pi }{4}}ln\left(cos\theta \right)d\theta but$$$${\int }_0^{\frac{\pi }{4}}cos(\theta -\frac{\pi }{4})d\theta ={\int }_0^{\frac{\pi }{4}}cos(\frac{\pi }{4}-\theta )d\theta {=}_{\frac{\pi }{4}-\theta =u}{\int }_{\frac{\pi }{4}}^{0}cos\left(u\right)(-du)$$$$={\int }_0^{\frac{\pi }{4}}cos\left(u\right)du\Rightarrow ★{\int }_0^1\frac{ln(1+x)}{1+x^2}dx=\frac{\pi }{8}ln\left(2\right)★$$

Commented by Necxx last updated on 22/Oct/18

$$thankyousomuch$$

Commented by Necxx last updated on 22/Oct/18

$$thankyousir$$

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