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SOLUTION 16:    Integrate 1xx+3 dx. Begin by rewriting the integrand. 1xx+3 dx=x22x+3 dx =(x2+2x)+3 dx (Now complete the square.) =(x2+2x+1)+3+1 dx =4(x+1)2 dx Now use the trig substitution x+1=2sinθ so that dx=2cosθ dθ Substitute into the original problem, replacing all forms of x, getting 4(x+1)2 dx=4(2sinθ)22cosθ dθ =244sin2θcosθ dθ =24(1sin2θ)cosθ dθ =24cos2θcosθ dθ =22cosθcosθ dθ =4cos2θ dθ (Recall that cos2θ=(1/2)(1+cos2θ).) =412(1+cos2θ) dθ =2(1+cos2θ) dθ =2(θ+12sin2θ)+C (Recall that sin2θ=2sinθcosθ.) =2(θ+12(2sinθcosθ))+C =2(θ+sinθcosθ)+C =2θ+2sinθcosθ+C ( We need to write our final answer in terms of x.

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Since x+1=2sinθ it follows that sinθ=x+12=oppositehypotenuse and θ=arcsinx+12 and from the Pythagorean Theorem that (adjacent)2+(opposite)2=(hypotenuse)2   (adjacent)2+(x+1)2=(2)2     adjacent=4(x+1)2=x22x+3   cosθ=adjacenthypotenuse=x22x+32.) =2arcsinx+12+2x+12x22x+32 =2arcsinx+12+x+12x22x+3

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