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  • Integrate the functions 3x upon 1 plus 2x raised to 4

 Q 5  Integrate the functions \frac{3x }{1+ 2 x ^ 4 }

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\frac{3x }{1+ 2 x ^ 4 }

 

Let {\sqrt{2}}x^2 = t 
\therefore2\sqrt{2}xdx =dt

The integration can be done as follows

\Rightarrow \int \frac{3x}{1+2x^4}= \frac{3}{2\sqrt{2}}\int \frac{dt}{1+t^2}
                                \\= \frac{3}{2\sqrt{2}}[\tan^{-1}t]+C\\ =\frac{3}{2\sqrt{2}}[\tan^{-1}(\sqrt{2}x^2)]+C

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manish

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