A few days ago
I need shown how to find the following antiderivatives. please show steps if u can. 1) 1/x^3 2)4/x?
3) 9/x^2 +4/11x
Please answer at least one if you can, if u can all 3 thatd be great.
Top 3 Answers
A few days ago
Favorite Answer
use the rule that
the anti derivative of x^n is [x^(n+1)] /(n+1)
except where n = -1
then you get
the anti derivative of 1/x = x^-1 is lnx
1)
1/x^3 = x^-3
so
the anti derivative is
[x^(-3+1)] / (-3+1)
=[ x^-2] /-2
= -1/2x^2
2)
4/x = 4*(1/x)
the anti derivative is
4lnx
3)
9/x^2 +4/11x
= 9x^-2 + (4/11)*(1/x)
the anti derivative is
[9x^(-2+1)] /(-2+1) + (4/11)*lnx
= [9x^-1] /-1 + (4/11)*lnx
= -9/x + (4lnx) /11
if you differentiate each of my answers you will get your questions back again
.
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A few days ago
Antiderivative means that if you differentiate your solution, you’ll get the problem question: For 1/x^3=x^(-3), add 1 to that (-3) to yield -2 as exponent of x and divide by the -2, which is (x^-2)/-2, so were you to differentiate, you’d cancel out the that -2 giving the problem x^-3. For 4/x, since the antiderivative of 1/x is ln x, the answer is 4 ln x, where x>0. Likewise, the solution to the third problem is 9(-x^-1)+(4/11) (x^2)/2=(-9x^-1)+2(x^2)/11, which if you differentiate will yield problem question.
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A few days ago
idem!!!
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