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AP Calculus—Integration Practice

AP Calculus integration by Idea:Ifu=f(x), thendu=f (x) have xdxx4+ 1u=x2=dx= 2xdx12 duu2+ 1=12tan 1u+C=12tan 1x2+CPractice Problems:1. x3 4 +x4dx2. dxxlnx3. (x+ 5)dx x+ 44. In each integral below, find the integernthat allows for an integration bysub-stitution. Then perform the integration .(a) xn 1 x4dx(b) xn 1 x4dx(there are two very natural choices forn).(c) xn1 +x10dx(there are two very natural choices forn).(d) x61 +xndx(e) xne x2dx(f) xne2x5dx(g) x5 1 xndx(h) x6 1 xndx(i) dxxnlnx(j) dxxn(lnx)7(k) xnsin(x6)dx(l) sinnxcosx 3 + sin4xdx(m) sin3xcosx 3 + by Parts:Basic Idea: udv=uv v du(Try to substituteuso thatdudxis simpler thanuand so thatvis no more complicatedthandv.)

Integration by Partial Fractions. Basic Idea: This is used to integrate rational functions. Namely, if R(x) = p(x) q(x) is a rational function, with p(x) and q(x) polynomials, then we can factor q(x) into a product of linear and irreducible quadratic factors, possibly with multiplicities. For

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