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Q: What is the antiderivative of logx?

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logx^2=2 2logx=2 logx=1 10^1=x x=10

log(100x) can be written as log100 + logx. This =2+logx

Using u-substitution (where u = sinx), you'll find the antiderivative to be 0.5*sin2x + C.

*First off if we assume this log to be base 10 next we can use the product rule (d/dx (3)*logx+d/dx(logx)*3) 1.derivative of a constant is zero so that gives us 0*logx as our first term (simplifies to zero) next we have to differentiate logx that gives us 3*(1/xln(10)) so that leaves 0logx+3*(1/xln(10)) simplify...... 3/xln(10)

(2/3)*x^(3/2)

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logx^3logx^2log14 is 3logx2logxlog14 this equals 6 log14 (logx)^2 So for example, if y=6log14(logx)^2 the log x = square root of (y/6(log14))

The antiderivative of 2x is x2.

logx^2=2 2logx=2 logx=1 10^1=x x=10

log(100x) can be written as log100 + logx. This =2+logx

y=logx y=10 logx= 10 10logx = 10log1 logx = log1 x = 1 //NajN

The derivative of logx, assuming base 10, is 1/(xln10).

-logx=21.1 logx=-21.1 e^-21.1=x

The antiderivative of a function which is equal to 0 everywhere is a function equal to 0 everywhere.

logx = 2 so x = 10logx = 102 = 100 ie x = 100.

35x2

Using u-substitution (where u = sinx), you'll find the antiderivative to be 0.5*sin2x + C.

I assume you mean -10x^4? In that case, antiderivative would be to add one to the exponent, then divide by the exponent. So -10x^5, then divide by 5. So the antiderivative is -2x^5.

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