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Worked Examples · Example 3

Q.Evaluate the following:

(a) ∫2x+3 dx\int \sqrt{2x+3}\,dx
(b) ∫e4−5x dx\int e^{4-5x}\,dx
(c) ∫(ax+b)2 dx\int (ax+b)^2\,dx
(d) ∫[ax+a−x]2dx\int \left[a^x+a^{-x}\right]^2 dx
CBSENCERTSubjective· 5mImportance★★★★★
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✓ Free question

Each uses the linear-substitution rule ∫f(px+q) dx=1pF(px+q)\int f(px+q)\,dx=\frac1p F(px+q); part (d) is expanded first.

∫(px+q)ndx=(px+q)n+1p(n+1)+C,∫epx+qdx=epx+qp+C,∫akxdx=akxklog⁡a+C.\int(px+q)^n dx=\frac{(px+q)^{n+1}}{p(n+1)}+C,\quad \int e^{px+q}dx=\frac{e^{px+q}}{p}+C,\quad \int a^{kx}dx=\frac{a^{kx}}{k\log a}+C.

Steps

  1. (a) p=2p=2, n=12n=\tfrac12:

∫(2x+3)1/2dx=(2x+3)3/22⋅32=13(2x+3)3/2+C.\int(2x+3)^{1/2}dx=\frac{(2x+3)^{3/2}}{2\cdot\frac32}=\frac{1}{3}(2x+3)^{3/2}+C.

  1. (b) p=−5p=-5:

∫e4−5xdx=e4−5x−5=−15e4−5x+C.\int e^{4-5x}dx=\frac{e^{4-5x}}{-5}=-\frac{1}{5}e^{4-5x}+C.

  1. (c) p=ap=a, n=2n=2:

∫(ax+b)2dx=(ax+b)3a⋅3=(ax+b)33a+C.\int(ax+b)^2dx=\frac{(ax+b)^3}{a\cdot3}=\frac{(ax+b)^3}{3a}+C.

  1. (d) Expand: (ax+a−x)2=a2x+2+a−2x(a^x+a^{-x})^2=a^{2x}+2+a^{-2x}.

∫ ⁣(a2x+2+a−2x)dx=a2x2log⁡a+2x+a−2x−2log⁡a=a2x−a−2x2log⁡a+2x+C.\int\!\big(a^{2x}+2+a^{-2x}\big)dx=\frac{a^{2x}}{2\log a}+2x+\frac{a^{-2x}}{-2\log a}=\frac{a^{2x}-a^{-2x}}{2\log a}+2x+C.

✓Final answer

  1. 13(2x+3)3/2+C\dfrac{1}{3}(2x+3)^{3/2}+C;
  2. −15e4−5x+C-\dfrac{1}{5}e^{4-5x}+C;
  3. (ax+b)33a+C\dfrac{(ax+b)^3}{3a}+C;
  4. a2x−a−2x2log⁡a+2x+C\dfrac{a^{2x}-a^{-2x}}{2\log a}+2x+C.

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