Fie \( f:[0,\infty)\to \mathbb{R},\ f(x)=\int_0^{\arctan x}\ln (1+\tan t){\rm d}t \).
i) Sa se calculeze \( f(0),\ f(1) \).
ii) Sa se calculeze \( f^\prime \).
iii) Calculati \( \int_0^1\frac{\ln (1+x)}{x^2+1}{\rm d}x \).
iv) Calculati \( \int_0^1xf(x){\rm d} x \).
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Beniamin Bogosel
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i) \( f(0)=\int_0^0\ln(1+\tan t)dt=0 \)
\( f(1)=\int_0^{\frac{\pi}{4}}\ln(1+\tan t)dt=\int_0^{\frac{\pi}{4}}\ln(1+\tan(\frac{\pi}{4}-t))dt=\int_0^{\frac{\pi}{4}}\ln2dt-f(1) \)
si de aici \( f(1)=\frac{\pi\ln2}{8} \)
ii) \( f^{\prime}(x)=\frac{\ln(1+x)}{1+x^2} \)
iii) \( \int_0^1\frac{\ln(1+x)}{1+x^2}dx=f(1)=\frac{\pi\ln2}{8} \)
iv) \( \int_0^1xf(x)dx=\frac{x^2}{2}f(x)|_0^1-\int_0^1\frac{x^2}{2}\cdot\frac{\ln(1+x)}{1+x^2}dx=\frac{f(1)}{2}-\frac{1}{2}\int_0^1\ln(1+x)dx+\frac{1}{2}\int_0^1f\prime(x)dx=f(1)-\frac{1}{2}x\ln(1+x)|_0^1+\frac{1}{2}\int_0^1\frac{x}{1+x}dx=\frac{\pi\ln2}{8}-\ln2+\frac{1}{2} \)
\( f(1)=\int_0^{\frac{\pi}{4}}\ln(1+\tan t)dt=\int_0^{\frac{\pi}{4}}\ln(1+\tan(\frac{\pi}{4}-t))dt=\int_0^{\frac{\pi}{4}}\ln2dt-f(1) \)
si de aici \( f(1)=\frac{\pi\ln2}{8} \)
ii) \( f^{\prime}(x)=\frac{\ln(1+x)}{1+x^2} \)
iii) \( \int_0^1\frac{\ln(1+x)}{1+x^2}dx=f(1)=\frac{\pi\ln2}{8} \)
iv) \( \int_0^1xf(x)dx=\frac{x^2}{2}f(x)|_0^1-\int_0^1\frac{x^2}{2}\cdot\frac{\ln(1+x)}{1+x^2}dx=\frac{f(1)}{2}-\frac{1}{2}\int_0^1\ln(1+x)dx+\frac{1}{2}\int_0^1f\prime(x)dx=f(1)-\frac{1}{2}x\ln(1+x)|_0^1+\frac{1}{2}\int_0^1\frac{x}{1+x}dx=\frac{\pi\ln2}{8}-\ln2+\frac{1}{2} \)