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Section 5.1 #44) Compute ddxln(2x2+1).
Solution: Compute ddxln(2x2+1)=12x2+1ddx[2x2+1]=4x2x2+1.

Section 5.1 #57) Compute ddxln(x+1x1).
Solution: A more difficult (but direct) method than I will use would be to directly compute ddxln(x+1x1)=1x+1x1ddx[x+1x1], which would then require a second chain rule involving the quotient rule. The easier method for this antiderivative is to rewrite the symbol using exponent 12 and use the property of logarithms to move that 12 to the front. After that, integrate: compute ddxln(x+1x1)=ddxln((x+1x1)12)=12ddxln(x+1x1)=121x+1x1ddxx+1x1=12x1x+1(x1)(x+1)(x1)2=121x+12x1=1x21=11x2.

Section 5.2 #8) Compute x25x3dx.
Solution: Let u=5x3 so that du=3x2dx. Express that as 13du=x2dx. Now compute x25x3dx=131udu=13log(u)+C=13log(5x3)+C. Section 5.2 #31) Compute cot(θ3)dθ.
Solution: Recall that by definition, cot(x)=cos(x)sin(x), and so cot(θ3)=cos(θ3)sin(θ3). Let u=sin(θ3) so that du=13cos(θ3)dx. Rearrange this formula to get 3du=cos(θ3)dθ. Now compute cot(θ3)dθ=cos(θ3)sin(θ3)dθ=31udu=3ln(|u|)+C=3ln(|sin(θ3)|)+C.

Section 5.2 #49) Compute 4053x+1dx.
Solution:

Section 5.2 #55) Compute 211cos(θ)θsin(θ)dθ.
Solution: Let u=θsin(θ) so that du=1cos(θ)dθ. If θ=1 then u=1sin(1) and if θ=2 then u=2sin(2). Now compute 211cos(θ)θsin(θ)dθ=2sin(2)1sin(1)1udu=ln(|u|)|2sin(2)1sin(1)=ln(|2sin(2)|)ln(|1sin(1)|).