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Worked Examples · Example 9

Q.Integrate the following:

(a) xe2xxe^{2x}
(b) log⁡x\log x
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✓ Free question

Both use integration by parts, ∫u dv=uv−∫v du\int u\,dv=uv-\int v\,du (ILATE choice of uu).

∫u dv=uv−∫v du.\int u\,dv=uv-\int v\,du.

Steps

  1. (a) ∫xe2xdx\int xe^{2x}dx: take u=x, dv=e2xdx⇒du=dx, v=12e2xu=x,\ dv=e^{2x}dx\Rightarrow du=dx,\ v=\tfrac12 e^{2x}.

=x⋅e2x2−∫e2x2 dx=xe2x2−e2x4+C=e2x4(2x−1)+C.=x\cdot\frac{e^{2x}}{2}-\int\frac{e^{2x}}{2}\,dx=\frac{xe^{2x}}{2}-\frac{e^{2x}}{4}+C=\frac{e^{2x}}{4}(2x-1)+C.

  1. (b) ∫log⁡x dx=∫1⋅log⁡x dx\int\log x\,dx=\int 1\cdot\log x\,dx: take u=log⁡x, dv=dx⇒du=1xdx, v=xu=\log x,\ dv=dx\Rightarrow du=\tfrac1x dx,\ v=x.

=xlog⁡x−∫x⋅1x dx=xlog⁡x−∫1 dx=xlog⁡x−x+C.=x\log x-\int x\cdot\frac1x\,dx=x\log x-\int 1\,dx=x\log x-x+C.

(Here log⁡\log is the natural logarithm, as standard in this chapter.)

✓Final answer

  1. e2x4(2x−1)+C\dfrac{e^{2x}}{4}(2x-1)+C;
  2. xlog⁡x−x+Cx\log x-x+C.

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