How do you integrate #8x(x²+1)³ dx #?

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Answer 1 · 2016-09-16T18:48:31

#(x^2+1)^4+C#.

Let #I=int8x(x^2+1)^3dx#

We use the Method of Substn.

We subst. #(x^2+1)=y," so that, "2xdx=dy#. Hence,

#I=4inty^3dy=4*y^4/4=y^4=(x^2+1)^4+C#.

Alternatively, without substn., we can find #I#.

Using #(a+b)^3=a^3+b^3+3ab(a+b)#, we expand #(x^2+1)^3#.

#:. I=int{8x(x^6+1+3*x^2*1(x^2+1)}dx#

#=8int(x^7+x+3x^5+3x^3)dx#

#=8{x^8/8+x^2/2+3*x^6/6+3*x^4/4}#

#=x^8+4x^2+4x^6+6x^4=x^8+4x^6+6x^4+4x^2+K#.

On the first hand, these two Answers may look different, but,

bearing in mind the binomial expansion of

#(p+1)^4=p^4+4p^3+6p^2+4p+1#, we find that,

#I=x^8+4x^6+6x^4+4x^2+K#,

#=x^8+4x^6+6x^4+4x^2+1+C'#, where, #C'=K-1#

#I=(x^2+1)^4+C', C'=K-1#

Enjoy Maths.!