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ludmilkaskok [199]
3 years ago
15

Please help ASAP!!!!!!!

Mathematics
1 answer:
vladimir2022 [97]3 years ago
7 0

Answer: c

Step-by-step explanation:

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Integration by parts will help here. Letting u=\arctan x and \mathrm dv=\mathrm dx, you end up with \mathrm du=\dfrac{\mathrm dx}{1+x^2} and v=x. Now

\displaystyle\int\arctan x\,\mathrm dx=uv-\int v\,\mathrm du
\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\int\frac x{1+x^2}\,\mathrm dx

For the remaining integral, setting y=1+x^2 gives \dfrac{\mathrm dy}2=x\,\mathrm dx, so

\displaystyle\int\frac x{1+x^2}\,\mathrm dx=\frac12\int\frac{\mathrm dy}y=\frac12\ln|y|+C=\frac12\ln(1+x^2)+C

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\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\frac12\ln(1+x^2)+C
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Natasha_Volkova [10]

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