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Question Number 107496 by mathdave last updated on 11/Aug/20

Answered by bemath last updated on 11/Aug/20

$$\nparallel \mathcal{B}e\mathcal{M}ath\nparallel$$$$=\int {(1+\tan {}^{2}x)}^2\sec {}^{2}xdx$$$$=\int {(1+u^2)}^{2}du[u=\tan x]$$$$=\int (1+2u^2+u^4)du$$$$=u+\frac{2}{3}{u}^3+\frac{1}{5}{u}^5+C$$$$=\tan x(1+\frac{2}{3}\tan {}^{2}x+\frac{1}{5}\tan {}^{4}x)+C$$

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