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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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I need to use the Pythagorean Theorem to find X. I don’t know how to do that!!
frutty [35]

Answer:

17) x=8

18) x=\sqrt{189} or x= 3\sqrt{21}

Step-by-step explanation:

So the rule is a^{2} +b^{2} =c^{2}, "c" being the hypothenuse, or the long line that is opposite to the right angle.

17) We know that both values of x are equal to each other, which makes everything 10x easier!

                                               x^{2} +x^{2} =(8\sqrt{2})  ^{2}

(by the way we know the x values are our a and b values because they are legs! the way I like to remember the legs is that they are connected to the right angle box, and therefore support the hypothenuse)

<em>simplify</em> (╥︣﹏╥)    

                                                2x^2=8^2(\sqrt{2} )^{2}  

                                                  2x^2=64(2)

                                                   2x^2=128    

                                                  x^{2}= 128/2  

                                                     x^{2} =64

                                                        x=8      

18) Just pretend that the flipped triangle doesn't exist. It's parallel to the other triangle with values on it, and basically servers no purpose other than being parallel to the sister triangle :)

Anyways, since we know the hypothenuse (15) but we don't know one of our leg values (x), we're going to change our equation a bit!

                                                    c^{2} - b^{2} = a^{2}

It doesn't matter if you put the one leg value in a or b, just as long as you stick to that same equation you started with the entire time!

                                                   15^{2} -6^{2} =x^{2}

                                                   225-36=x^{2}  

                                                     x^{2} =189

                                               x=\sqrt{189}=3\sqrt{21}

        The more you do these, the easier they'll get, so don't worry!

                           

4 0
3 years ago
What is the sum of numbers as a product of their GCF 45+60
Eduardwww [97]
Find the prime factorization

45=3*3*5
60=2*2*3*5
GCF=3*5=15

45=3*15
60=4*15

remember
ab+ac=a(b+c) so
45+60=15(3)+15(4)=15(3+4)=15(7)=105
5 0
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Answer:

6

4.75÷.75=6

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