Integration using sub. Q (1 Viewer)

flon · Feb 20, 2009

(indefinite integral) 5x(x+4)^1/2 dx

using the sub. u = x+4

Thanks in advance.

Aerath · Feb 20, 2009

vds700 · Feb 20, 2009

$\int 5x\sqrt{x+4}dx\\$let$u=x+4\Rightarrow \frac{du}{dx}=1\Rightarrow du=dx\\\int5x\sqrt{x+4}dx=5\int (u+4)u^{\frac{1}{2}}du\\=5\int (u^{\frac{3}{2}}+4u^{\frac{1}{2}})du\\=5[\frac{2u^{5/2}}{5}+4.\frac{2u^{3/2}}{3}]+c\\=2u^{\frac{5}{2}}+\frac{40u^{\frac{3}{2}}}{3}+c$ Now Just put back in terms of x, sorry, i cbf

vds700 · Feb 20, 2009

Aerath said:

i think u misread the question its x + 4, not x² + 4

flon · Feb 20, 2009

vds700 said:
$\int 5x\sqrt{x+4}dx\\$let$u=x+4\Rightarrow \frac{du}{dx}=1\Rightarrow du=dx\\\int5x\sqrt{x+4}dx=5\int (u+4)u^{\frac{1}{2}}du\\=5\int (u^{\frac{3}{2}}+4u^{\frac{1}{2}})du\\=5[\frac{2u^{5/2}}{5}+4.\frac{2u^{3/2}}{3}]+c\\=2u^{\frac{5}{2}}+\frac{40u^{\frac{3}{2}}}{3}+c$ Now Just put back in terms of x, sorry, i cbf

thanks got it

vds700 · Feb 20, 2009

flon said:
thanks got it

just realised i made a mistake, change the (u +4)'s to (u - 4)- sorry

Aerath · Feb 21, 2009

flon said:
thanks got it

Hmph, yeah, my bad. =P

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Integration using sub. Q (1 Viewer)

flon

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vds700

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flon

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vds700

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Aerath

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