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Evaluate the following : ∫ln(x) dx

Integrate by parts:u = ln(x) u' = 1/x v' = 1 v = xBy parts formula: uv - ∫u'v dx Therefore we have: x ln(x) - ∫x 1/x dx = x ln(x) - ∫1 dx = x ln(x) - x (+c)
SM
Answered by Saaqib M. • Maths tutor
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What is the chain rule and how is it used?

The chain rule is used to find the derivative of an expression in the form h(f(x)) where you have a function in terms of a function of x for example: h(f(x)) = 2(3x+1)^3 where f(x) = 3x+1 In order to use the...
JB
Answered by Joseph B. • Maths tutor
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Use the addition formulas: sin(x+y)=sin(x)*cos(y)+sin(y)*cos(x), cos(x+y)=cos(x)*cos(y)-sin(x)*sin(y) to derive sin(2x), cos(2x), sin(x)+sin(y).

sin(2x)=sin(x+x) =sin(x)cos(x)+cos(x)sin(x) =2*sin(x)*cos(x); cos(2x)=cos(x+x) =cos(x)cos(x)-sin(x)sin(x) =[cos(x)] 2 -[sin(x)] 2 Now let x+y=a and x-y=b, then x=(a+b)/2 and y=(a-b)/2. Then sin(a)=sin[(a+b)/...
IS
Answered by Ioana S. • Maths tutor
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A curve has the equation, 6x^2 +3xy−y^2 +6=0 and passes through the point A (-5, 10). Find the equation of the normal to the curve at A.

Use implicit differentiation on original equation- 12x + 3x(dy/dx) + 3y - 2y(dy/dx) = 0 dy/dx= -12x -3y/(3x-2y) at A, x= -5 and y= 10 therefore, dy/dx=-6/7 To find the normal of the curve, use the negative r...
CG
Answered by Chantal G. • Maths tutor
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A curve has the equation y = (x^2 - 5)e^(x^2). Find the x-coordinates of the stationary points of the curve.

This requires the chain rule and the product rule to be used to differentiate the function. The substitution u = x 2 can be used to make this easier. Using this, du/dx = 2x and y = (u-5)e u . Using the produ...
OJ
Answered by Oliver J. • Maths tutor
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