How did they do this?

  • #1

Main Question or Discussion Point

How did they did they do the second line? Any help appreciated, thanks in advance.
 

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  • #2
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You know the rules of integration right? Imagine differentiating [itex]x^3[/itex]. You get [itex]3x^2[/itex].
Now reverse the differentiation.

So integrating [itex]3x^2[/itex] you first increase the power and then divide by the new power. So...

1) Raise the power: [itex]3x^3[/itex]
2) Divide by the new power: [itex]\frac{3}{3}x^3[/itex] = [itex]x^3[/itex]

That's all that's happening on the second line.
 
  • #3
575
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##\int 3x^\frac{3}{4}+7x^{-5}+\frac{1}{6}x^{-\frac{1}{2}}\ dx=\int 3x^\frac{3}{4}\ dx+\int 7x^{-5}\ dx+\int \frac{1}{6}x^{-\frac{1}{2}}\ dx## using the sum rule for integrals


##\int 3x^\frac{3}{4}\ dx+\int 7x^{-5}\ dx+\int \frac{1}{6}x^{-\frac{1}{2}}\ dx=3\int x^\frac{3}{4}\ dx+7\int x^{-5}\ dx+\frac{1}{6}\int x^{-\frac{1}{2}}\ dx## using the constant multiple rule for integrals


##3\int x^\frac{3}{4}\ dx+7\int x^{-5}\ dx+\frac{1}{6}\int x^{-\frac{1}{2}}\ dx=3\cdot\frac{1}{\frac{3}{4}+1} x^{\frac{3}{4}+1}+7\cdot\frac{1}{-5+1}x^{-5+1}+\frac{1}{6}\cdot\frac{1}{-\frac{1}{2}+1} x^{-\frac{1}{2}+1}+c## using the integral formula ##\int x^\alpha\ dx=\frac{1}{a+1}x^{\alpha+1}+c## when ##\alpha\neq-1##
 
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  • #4
Profuse thanks dear sir.
 

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