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help needed (1 Viewer)

Thread starter Abhiram Yanamandra
Start date Jul 2, 2019

Abhiram Yanamandra

New Member

Jul 2, 2019

how to do this qn. My working out gave a anwer i think is wrong.

$\int ^{\frac{1}{3}} _0 \frac {e^{3x} }{e^{6x} + 1} dx =\int ^{\frac{1}{3}} _ 0 \frac {e^{3x}}{e^{6x} + 1} dx =\frac {1}{3} \int ^- _- \frac {1}{1 + (e^{3x})^2} d(e^{3x}) \\ \\ = \frac {1}{3} \left [ tan^{-1} e^{3x} \right ]^\frac{1}{3} _0 = \frac {1}{3}[tan^{-1} e^1 - tan ^{-1} e^0 ] = \frac {1}{3}[tan^{-1} e - \frac {\pi}{4}]$