Question Number 7482 by kishanjith last updated on 31/Aug/16 $$\int\frac{\left(\mathrm{1}+{x}\right)^{−\mathrm{2}/\mathrm{3}} }{\left(\mathrm{1}+{x}\right)}{dx} \\ $$ Answered by sandy_suhendra last updated on 31/Aug/16 $$=\int\left(\mathrm{1}−{x}\right)^{\frac{−\mathrm{5}}{\mathrm{3}}} {dx} \\ $$$${let}\:\:\:{U}=\mathrm{1}−{x}\: \\…
Question Number 138554 by cherokeesay last updated on 14/Apr/21 Answered by mr W last updated on 14/Apr/21 Commented by mr W last updated on 14/Apr/21…
Question Number 73017 by Rio Michael last updated on 05/Nov/19 $${find}\: \\ $$$$\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\:\:{x}\sqrt{{x}^{\mathrm{2}} \:+\:\mathrm{1}}\: \\ $$ Commented by mathmax by abdo last updated on…
Question Number 138548 by I want to learn more last updated on 14/Apr/21 Answered by nimnim last updated on 14/Apr/21 $${Let}\:{the}\:{centre}\:{of}\:{the}\:{circle}\:{be}\:{C}\left({h},{k}\right) \\ $$$${Since},\:{it}\:{touches}\:{the}\:{x}−{axis}\:{at}\:{P}\left(−\mathrm{3},\mathrm{0}\right), \\ $$$$\:{CP}\bot\:{x}−{axis}\:{and}\:{h}=−\mathrm{3}…
Question Number 7478 by Yozzia last updated on 31/Aug/16 $${Find}\:{the}\:{value}\:{of}\:\:{x}^{\mathrm{3}} \left({x}^{\mathrm{3}} −\mathrm{18}\right)\:\:{if}\:\:{x}\left({x}−\mathrm{3}\right)=−\mathrm{1}. \\ $$ Answered by Rasheed Soomro last updated on 31/Aug/16 $$\:\:\:{x}^{\mathrm{3}} \left({x}^{\mathrm{3}} −\mathrm{18}\right)\:\:{if}\:\:{x}\left({x}−\mathrm{3}\right)=−\mathrm{1}…
Question Number 73012 by vishalbhardwaj last updated on 05/Nov/19 $$\int\sqrt{{tan}\mathrm{x}}\:{d}\mathrm{x} \\ $$ Answered by mind is power last updated on 05/Nov/19 $$\mathrm{u}=\sqrt{\mathrm{tan}\left(\mathrm{x}\right)} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{arctan}\left(\mathrm{u}^{\mathrm{2}} \right)\Rightarrow\mathrm{dx}=\frac{\mathrm{2u}}{\mathrm{1}+\mathrm{u}^{\mathrm{4}}…
Question Number 7474 by Tawakalitu. last updated on 31/Aug/16 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 138546 by mnjuly1970 last updated on 14/Apr/21 Answered by ajfour last updated on 14/Apr/21 $$\left({i}\right)\:\:{a}^{\mathrm{3}} ={p},\:{b}^{\mathrm{3}} ={q},\:{c}^{\mathrm{3}} ={r} \\ $$$$\:{a}+{b}+{c}=\:\boldsymbol{{v}}\:=\sqrt[{\mathrm{3}}]{{p}}+\sqrt[{\mathrm{3}}]{{q}}+\sqrt[{\mathrm{3}}]{{r}} \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{3}} ={v}^{\mathrm{3}}…
Question Number 7472 by FilupSmith last updated on 31/Aug/16 $$\mathrm{Show}\:\mathrm{why}: \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:{x}^{{x}} \:=\:\mathrm{0} \\ $$ Commented by FilupSmith last updated on 31/Aug/16 $$=\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:{e}^{{x}\mathrm{ln}\left({x}\right)}…
Question Number 138536 by Tinku Tara last updated on 14/Apr/21 $$\mathrm{This}\:\mathrm{app}\:\mathrm{is}\:\mathrm{now}\:\mathrm{free}\:\mathrm{as}\:\mathrm{several}\:\mathrm{users} \\ $$$$\mathrm{do}\:\mathrm{not}\:\mathrm{have}\:\mathrm{cards}\:\mathrm{and}\:\mathrm{are}\:\mathrm{not}\:\mathrm{able} \\ $$$$\mathrm{to}\:\mathrm{upgrade}.\:\mathrm{Manual}\:\mathrm{entry}\:\mathrm{in}\:\mathrm{backend} \\ $$$$\mathrm{system}\:\mathrm{had}\:\mathrm{put}\:\mathrm{lot}\:\mathrm{of}\:\mathrm{overhead}. \\ $$$$\mathrm{Hence}\:\mathrm{we}\:\mathrm{have}\:\mathrm{decide}\:\mathrm{to}\:\mathrm{make}\:\mathrm{this} \\ $$$$\mathrm{app}\:\mathrm{free}. \\ $$$$\mathrm{We}\:\mathrm{issued}\:\mathrm{several}\:\mathrm{refunds}\:\mathrm{for}\:\mathrm{users} \\ $$$$\mathrm{who}\:\mathrm{had}\:\mathrm{bought}.\:\mathrm{However}\:\mathrm{Google}…