Question-113323

Question Number 113323 by AbhishekBasnet last updated on 12/Sep/20

Answered by bemath last updated on 12/Sep/20

let a+bx = u , x = ((u−a)/b) ∫ (((u−a)/b))u^3 ((du/b)) = (1/b^2 )∫ (u^4 −au^3 )du =(1/b^2 )((u^5 /5)−((au^4 )/4)) + c =(u^4 /(20b^2 ))(4u−5a) + c =(((a+bx)^4 )/(20b^2 ))(4bx−a)+c

$$\mathrm{let}\:\mathrm{a}+\mathrm{bx}\:=\:\mathrm{u}\:,\:\mathrm{x}\:=\:\frac{\mathrm{u}−\mathrm{a}}{\mathrm{b}} \\ $$$$\int\:\left(\frac{\mathrm{u}−\mathrm{a}}{\mathrm{b}}\right)\mathrm{u}^{\mathrm{3}} \left(\frac{\mathrm{du}}{\mathrm{b}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{2}} }\int\:\left(\mathrm{u}^{\mathrm{4}} −\mathrm{au}^{\mathrm{3}} \right)\mathrm{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{2}} }\left(\frac{\mathrm{u}^{\mathrm{5}} }{\mathrm{5}}−\frac{\mathrm{au}^{\mathrm{4}} }{\mathrm{4}}\right)\:+\:\mathrm{c} \\ $$$$=\frac{\mathrm{u}^{\mathrm{4}} }{\mathrm{20b}^{\mathrm{2}} }\left(\mathrm{4u}−\mathrm{5a}\right)\:+\:\mathrm{c} \\ $$$$=\frac{\left(\mathrm{a}+\mathrm{bx}\right)^{\mathrm{4}} }{\mathrm{20b}^{\mathrm{2}} }\left(\mathrm{4bx}−\mathrm{a}\right)+\mathrm{c} \\ $$

Answered by bemath last updated on 12/Sep/20

by parts { ((u=x→du=dx)),((v=∫(a+bx)^3 dx=(1/(4b))(a+bx)^4 )) :} I= ((x(a+bx)^4 )/(4b))−∫ (1/(4b))(a+bx)^4 dx I= ((x(a+bx)^4 )/(4b))−(((a+bx)^5 )/(20b^2 ))+c

$$\mathrm{by}\:\mathrm{parts}\:\begin{cases}{\mathrm{u}=\mathrm{x}\rightarrow\mathrm{du}=\mathrm{dx}}\\{\mathrm{v}=\int\left(\mathrm{a}+\mathrm{bx}\right)^{\mathrm{3}} \mathrm{dx}=\frac{\mathrm{1}}{\mathrm{4b}}\left(\mathrm{a}+\mathrm{bx}\right)^{\mathrm{4}} }\end{cases} \\ $$$$\mathrm{I}=\:\frac{\mathrm{x}\left(\mathrm{a}+\mathrm{bx}\right)^{\mathrm{4}} }{\mathrm{4b}}−\int\:\frac{\mathrm{1}}{\mathrm{4b}}\left(\mathrm{a}+\mathrm{bx}\right)^{\mathrm{4}} \mathrm{dx} \\ $$$$\mathrm{I}=\:\frac{\mathrm{x}\left(\mathrm{a}+\mathrm{bx}\right)^{\mathrm{4}} }{\mathrm{4b}}−\frac{\left(\mathrm{a}+\mathrm{bx}\right)^{\mathrm{5}} }{\mathrm{20b}^{\mathrm{2}} }+\mathrm{c} \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply