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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+1x−1).
Solution: A more difficult (but direct) method than I will use would be to directly compute
ddxln(√x+1x−1)=1√x+1x−1ddx[√x+1x−1],
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+1x−1)=ddxln((x+1x−1)12)=12ddxln(x+1x−1)=121x+1x−1ddxx+1x−1=12x−1x+1(x−1)−(x+1)(x−1)2=121x+1−2x−1=−1x2−1=11−x2.◼
Section 5.2 #8) Compute ∫x25−x3dx.
Solution: Let u=5−x3 so that du=−3x2dx. Express that as −13du=x2dx. Now compute
∫x25−x3dx=−13∫1udu=−13log(u)+C=−13log(5−x3)+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=3∫1udu=3ln(|u|)+C=3ln(|sin(θ3)|)+C.◼
Section 5.2 #49) Compute ∫4053x+1dx.
Solution:
Section 5.2 #55) Compute ∫211−cos(θ)θ−sin(θ)dθ.
Solution: Let u=θ−sin(θ) so that du=1−cos(θ)dθ. If θ=1 then u=1−sin(1) and if θ=2 then u=2−sin(2). Now compute
∫211−cos(θ)θ−sin(θ)dθ=∫2−sin(2)1−sin(1)1udu=ln(|u|)|2−sin(2)1−sin(1)=ln(|2−sin(2)|)−ln(|1−sin(1)|).◼