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Question Number 81657 by zainal tanjung last updated on 14/Feb/20

$$\:\:\underset{\mathrm{0}} {\overset{\mathrm{3}} {\int}}\frac{\mathrm{x}+\mathrm{1}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2x}\right)^{\mathrm{15}} }=.... \\$$

Commented by Tony Lin last updated on 14/Feb/20

$$\int\:\frac{{x}+\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}\right)^{\mathrm{15}} }{dx} \\$$$$\Rightarrow{let}\:{u}={x}^{\mathrm{2}} +\mathrm{2}{x},\:{then}\:\frac{{du}}{{dx}}=\mathrm{2}\left({x}+\mathrm{1}\right) \\$$$$\int\frac{\mathrm{1}}{\mathrm{2}{u}^{\mathrm{15}} }{du} \\$$$$=−\frac{\mathrm{1}}{\mathrm{28}{u}^{\mathrm{14}} }+{c} \\$$$$=−\frac{\mathrm{1}}{\mathrm{28}\left({x}^{\mathrm{2}} +\mathrm{2}{x}\right)}+{c} \\$$$$\int_{\mathrm{0}} ^{\mathrm{3}} \:\frac{{x}+\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}\right)^{\mathrm{15}} }{dx} \\$$$$=\left[−\frac{\mathrm{1}}{\mathrm{28}\left({x}^{\mathrm{2}} +\mathrm{2}{x}\right)}\right]_{\mathrm{0}} ^{\mathrm{3}} \\$$$$\Rightarrow{divergent} \\$$

Commented by zainal tanjung last updated on 14/Feb/20

$$\mathrm{Thanks}\:\mathrm{Sir}! \\$$