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Levart [38]
3 years ago
15

How do you integrate arctan(x dx? i think that if you simplify the integral you get:?

Mathematics
1 answer:
Temka [501]3 years ago
6 0
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

Putting everything together, you end up with

\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\frac12\ln(1+x^2)+C
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Answer:

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Step-by-step explanation:

Parallel lines share the same slope.  Given y = -6x – 1, we know that the equation of this new line has the form y = -6x – C.  The coordinates of the point (1, 4) determine the value of C:

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