How we Integrate 1/ ((x^10) +x)?

Question

1218 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-22T10:17:09

# ln |x|-1/9ln|1+x^9|+C#

#1/(x+x^10) = 1/(x(1+x^9)) = 1/x-x^8/(1+x^9)#

Thus

#int 1/(x+x^10) dx= int(1/x-x^8/(1+x^9))dx#
#qquad = int dx/x - int 1/9 {d(1+x^9)}/{1+x^9} #
#qquad = ln |x|-1/9ln|1+x^9|+C#

Answer 2 · 2018-03-22T13:40:35

#int dx/(x^10+x) = 1/9ln abs (x^9/(x^9+1))+C#

Evaluate:

#int dx/(x^10+x) = int dx/(x(x^9+1)) = 1/9 int (9x^8dx)/(x^9(x^9+1))#

Substitute now:

#u = x^9#, #du = 9x^8dx#

#int dx/(x^10+x) = 1/9int (du)/(u(u+1))#

Decompose in partial fractions:

#1/(u(u+1)) = A/u +B/(u+1)#

#1/(u(u+1)) = (A(u+1) +Bu) / (u(u+1))#

#1/(u(u+1)) = ((A +B)u+A) / (u(u+1))#

So, #A=1#, #B=-1# and:

#1/(u(u+1)) = 1/u -1/(u+1)#

#int dx/(x^10+x) = 1/9(int (du)/u -int (du)/(u+1))#

#int dx/(x^10+x) = 1/9(ln absu - ln abs(u+1))+C#

#int dx/(x^10+x) = 1/9ln abs (x^9/(x^9+1))+C#