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Question Number 152110 by peter frank last updated on 25/Aug/21

∫a^(mx) b^(nx) dx

$$\int\mathrm{a}^{\mathrm{mx}} \mathrm{b}^{\mathrm{nx}} \mathrm{dx} \\ $$

Answered by peter frank last updated on 25/Aug/21

∫(a^m b^n )^x   ......

$$\int\left(\mathrm{a}^{\mathrm{m}} \mathrm{b}^{\mathrm{n}} \right)^{\mathrm{x}} \\ $$$$...... \\ $$

Answered by Olaf_Thorendsen last updated on 26/Aug/21

F(x) = ∫a^(mx) b^(nx)  dx  F(x) = ∫e^((mlna+nlnb)x)  dx  F(x) = (e^((mlna+nlnb)x) /(mlna+nlnb))+C  F(x) = ((a^(mx) b^(nx) )/(ln(a^m b^n )))+C

$$\mathrm{F}\left({x}\right)\:=\:\int{a}^{{mx}} {b}^{{nx}} \:{dx} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\int{e}^{\left({m}\mathrm{ln}{a}+{n}\mathrm{ln}{b}\right){x}} \:{dx} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\frac{{e}^{\left({m}\mathrm{ln}{a}+{n}\mathrm{ln}{b}\right){x}} }{{m}\mathrm{ln}{a}+{n}\mathrm{ln}{b}}+\mathrm{C} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\frac{{a}^{{mx}} {b}^{{nx}} }{\mathrm{ln}\left({a}^{{m}} {b}^{{n}} \right)}+\mathrm{C} \\ $$

Commented by peter frank last updated on 26/Aug/21

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

Answered by puissant last updated on 26/Aug/21

K=∫a^(mx) b^(nx) dx   =a^(mx) ∫b^(nx) dx−∫((d/dx)(a^(mx) ).∫b^(nx) dx)  =a^(mx) (b^(nx) /(nlnb))−∫a^(mx) ×mlna(b^(nx) /(nlnb))dx  =((a^(mx) b^(nx) )/(nlnb))−((mlna)/(nlnb))∫a^(mx) b^(nx) dx  ⇒ K=((mlna)/(nlnb))×K=((a^(mx) b^(nx) )/(nlnb))  ⇒ K(((nlnb+mlna)/(nlnb)))=((a^(mx) b^(nx) )/(nlnb))    ⇒ ∴∵ K=((a^(mx) b^(nx) )/(mlna+nlnb))+C..

$${K}=\int{a}^{{mx}} {b}^{{nx}} {dx}\: \\ $$$$={a}^{{mx}} \int{b}^{{nx}} {dx}−\int\left(\frac{{d}}{{dx}}\left({a}^{{mx}} \right).\int{b}^{{nx}} {dx}\right) \\ $$$$={a}^{{mx}} \frac{{b}^{{nx}} }{{nlnb}}−\int{a}^{{mx}} ×{mlna}\frac{{b}^{{nx}} }{{nlnb}}{dx} \\ $$$$=\frac{{a}^{{mx}} {b}^{{nx}} }{{nlnb}}−\frac{{mlna}}{{nlnb}}\int{a}^{{mx}} {b}^{{nx}} {dx} \\ $$$$\Rightarrow\:{K}=\frac{{mlna}}{{nlnb}}×{K}=\frac{{a}^{{mx}} {b}^{{nx}} }{{nlnb}} \\ $$$$\Rightarrow\:{K}\left(\frac{{nlnb}+{mlna}}{{nlnb}}\right)=\frac{{a}^{{mx}} {b}^{{nx}} }{{nlnb}} \\ $$$$ \\ $$$$\Rightarrow\:\therefore\because\:{K}=\frac{{a}^{{mx}} {b}^{{nx}} }{{mlna}+{nlnb}}+{C}.. \\ $$

Commented by peter frank last updated on 26/Aug/21

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

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