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Integration x^2 (xsec^2x+tanx)/(xtanx+1)dx equals1. -x^2/(xtanx+1)+c2. 2log|xsinx+cosx|+c3. --×^2/(xtanx+1)+2log|xsinx+cosx|+c

Integration x^2 (xsec^2x+tanx)/(xtanx+1)dx equals1. -x^2/(xtanx+1)+c2. 2log|xsinx+cosx|+c3. --×^2/(xtanx+1)+2log|xsinx+cosx|+c

Grade:12

2 Answers

jagdish singh singh
173 Points
7 years ago
\hspace{-0.7 cm}$Let $I = \int x^2\cdot \left(\frac{x\sec^2 x+\tan x}{x\tan x+1}\right)dx,$ Using $\bf{I.B.P}$ and\\\\\\ Put $x\tan x+1 = t\;,$ Then $(x\sec^2 x+\tan x)dx = dt$\\\\\\ $I = x^2 \int \frac{1}{t}dt-2\int \left(x\int\frac{1}{t}dt\right)dx = \frac{x^2}{x\tan x+1}-2\int\frac{x}{x\tan x+1}dx$\\\\\\ So $I = \frac{x^2}{x\tan x+1}-2\int\frac{x\cos x}{x\sin x+\cos x}dx$
jagdish singh singh
173 Points
7 years ago
Above process is wrong , May be you mean \displaystyle \int x^2 \cdot \frac{x\sec^2 x+\tan x}{(x\tan x+1)^2}dx$

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