Mastering Integration: Common Mistakes in Homework | Correct Solutions

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Homework Statement



\int \frac{dx}{(1+4x^2)^{3/2}}

The Attempt at a Solution



\int \frac{dx}{(1+4x^2)^{3/2}}

Let x = \frac{1}{4}tan(u), dx = \frac{1}{4}sec^2(u)du

\frac{1}{4}\int{\frac{sec^2(u)du}{(sec^2(u))^{3/2}}}

\frac{1}{4}\int{\frac{1}{sec(u)}}du

\frac{1}{4}\int{cos(u)}du

\frac{1}{4}sin(u) + CDrawing a triangle with angle u:tan(u) = 4x

Therefore:

Opposite = 4x

Adjacent = 1

Hypotenuse = \sqrt{1^2 + (4x)^2} = \sqrt{1 + 16x^2}

sin(u) = \frac{4x}{\sqrt{1 + 16x^2}}\frac{1}{4}sin(u) + C

\frac{1}{4}(\frac{4x}{\sqrt{1 + 16x^2}}) + C

\frac{x}{\sqrt{1 + 16x^2}} + C

I personally can't see what I did wrong here, but this is not the correct answer :( Both my math book and Mathematica say that the correct answer should actually be

\frac{x}{\sqrt{1 + 4x^2}} + C

Any help is greatly appreciated :)
 
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Nevermind, I got it.

The substitution I used, x = \frac{1}{4}tan(u) does not produce the result I wanted because the \frac{1}{4} would be squared also. Instead, I made the substitution x = \frac{1}{2}tan(u) and that led me to the correct answer.
 
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