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Section 5.4 #49: Calculate ddxex+1ex1.
Solution: Using the quotient rule, compute ddxex+1ex1=(ex1)(ex)(ex+1)(ex)(ex1)2=ex[(ex1)(ex+1)](ex1)2=2ex(ex1)2. Section 5.4 #52: Calculate ddxe2xtan(2x).
Solution: Recall that ddxtan(x)=sec2(x). Therefore use the product rule to compute ddxe2xtan(2x)=ddx[e2x]tan(2x)+e2xddxtan(2x)=2e2xtan(2x)+2e2xsec2(2x). Section 5.4 #106: Calculate e2x+2ex+1exdx.
Solution: You could make the substitution u=ex, but it is easier to use algebra to simplify the integrand: e2x+2ex+1ex=ex+2+ex, so we may write e2x+2ex+1exdx=ex+2+exdx=exdx+2dx+exdx. The first two of these integrals are straightforward, and for the third one we make the substitution u=x so that du=dx and we compute e2x+2ex+1exdx=exdx+2dx+exdx=ex+2xeudu=ex+2xeu+C=ex+2xex+C.

Section 5.4 #107: Calculate extan(ex)dx.
Solution: First we recall how to compute tan(x)dx: since tan(x)=sin(x)cos(x) we make the substitution u=cos(x) so that du=sin(x)dx and we see tan(x)dx=sin(x)cos(x)dx=1udu=log(u)+C=log(cos(x))+C. Now we may proceed with the original problem. To do it, make the subtitution w=ex so that dw=exdx (note: I'm using w instead of u since I already used u above). Now compute extan(ex)dx=tan(w)dw=[log(cos(w))]+C=log(cos(ex))+C.

Section 5.5 #43: Calculate ddtt22t.
Solution: Recall that we may write 2t=elog(2t)=etlog(2). Now use the product rule to compute ddtt22t=ddt[t2]2t+t2ddtetlog(2)=(2t)(2t)+(t2)(log(2)etlog(2))=t2t[2+tlog(2)]. Section 5.5 #81: Calculate 105x3xdx.
Solution: Recall how to compute axdx for any a>0: first rewrite ax=elog(ax)=exlog(a) and use the u-substitution u=xlog(a) to get 1log(a)du=dx. Then, axdx=exlog(a)dx=1log(a)eudu=1log(a)eu+C=1log(a)exlog(a)+C=1log(a)ax+C. Applying this formula for a=5 and for a=3 allows us to compute 105x3xdx=105xdx103xdx=[1log(5)5x|10[1log(3)3x|10=51log(5)31log(3)=4log(5)2log(3).