Mathematics: A. one way to integrate ∫x√x+1 dx is to use integration by parts. do so to find the antiderivative. b. another way to evaluate the integral in part a is by using the substitution u=x+1. do so to find the antiderivative. c. compare the results from the two methods. if they do not look the same, explain how this can happen. discuss the advantages and disadvantages of each method.

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A. one way to integrate ∫x√x+1 dx is to use integration

gracieorman4

15.01.2022

By parts:

$u=x\implies\mathrm du=\mathrm dx$

$\mathrm dv=\sqrt{x+1}\,\mathrm dx\implies v=\dfrac23(x+1)^{3/2}$

$\displaystyle\int x\sqrt{x+1}\,\mathrm dx=\frac23x(x+1)^{3/2}-\frac23\int(x+1)^{3/2}\,\mathrm dx$

$\displaystyle\int x\sqrt{x+1}\,\mathrm dx=\frac23x(x+1)^{3/2}-\frac4{15}(x+1)^{5/2}+C$

$\displaystyle\int x\sqrt{x+1}\,\mathrm dx=\frac1{15}(x+1)^{3/2}\left(10x-4(x+1)\right)+C$

$\displaystyle\int x\sqrt{x+1}\,\mathrm dx=\frac2{15}(x+1)^{3/2}\left(3x-2)\right)+C$

By substitution:

$u=x+1\implies\mathrm du=\mathrm dx$

$\displaystyle\int x\sqrt{x+1}\,\mathrm dx=\int(u-1)\sqrt u\,\mathrm du=\int u^{3/2}-u^{1/2}\,\mathrm du=\frac25u^{5/2}-\frac23u^{3/2}+C$

$\displaystyle\int x\sqrt{x+1}\,\mathrm dx=\frac25(x+1)^{5/2}-\frac23(x+1)^{3/2}+C$

$\displaystyle\int x\sqrt{x+1}\,\mathrm dx=\frac2{15}(x+1)^{3/2}\left(3(x+1)-5\right)+C$

$\displaystyle\int x\sqrt{x+1}\,\mathrm dx=\frac2{15}(x+1)^{3/2}\left(3x-2\right)+C$

As you can see, no difference (as long as you simplify everything in a consistent way).

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