Evaluate: #int_1^2 1/(x^3+x) dx# ?

4 Answers
Aug 16, 2017

Use partial fractions.

Explanation:

#1/(x(x^2+1)) = A/x+(Bx+C)/(x^2+1)#

#Ax^2+A+Bx^2+Cx = 1#

#A = 1#, #C = 0# and #B - -1# so

#int_1^2 1/(x(x^2+1)) dx = int_1^2 (1/x-x/(x^2+1)) dx#

# = [lnx-1/2ln(x^2+1)]_1^2#

# = (ln2-1/2ln5)-(0-1/2ln2)#

# = 3/2ln2-1/2ln5 = 1/2ln(8/5)#

(E) None of the above.

Note

Typically we think about trigonometric substitution if we have a quadratic.
So the #x(x^2+1)# in the denominator may point us in that direction. But, as you say, it's not clear how to use it. So try something else.

Use partial fractions to get rid of the cubic.

It then turns out that we don't need trigonometric substitution.

Added Note
See the answer by Mason M. for a solution by trigonometric substitution.

Aug 16, 2017

Integral #= 1/2ln(8/5)#

Explanation:

#I = int_1^2 1/(x^3+x) dx#

#= int_1^2 1/(x(x^2+1)) dx#

Solving the indefinite integral.

#int 1/(x(x^2+1)) dx = int(1/x-x/(x^2+1)) dx#
[Partial fractions]

#= int(1/x)dx - intx/(x^2+1)dx#
[Linearity]

Solving #int1/xdx = ln(absx)#

Solving #intx/(x^2+1)dx#

Let #u=x^2+1 -> dx =1/(2x) du#

#intx/(x^2+1)dx = int 1/(2u) du = 1/2 ln(absu)#

Undo substitution

#= 1/2 ln(abs(x^2+1))#

#:. int 1/(x(x^2+1)) dx = ln(absx) - 1/2 ln(abs(x^2+1))#

Applying bounds

#int_1^2 1/(x(x^2+1)) dx =[ln(absx) - 1/2 ln(abs(x^2+1))]_1^2#

#= (ln2-1/2ln5)-(0-1/2ln2)#

#=ln2+1/2ln2-1/2ln5#

#=3/2ln2-1/2ln5#

#=1/2(3ln2-ln5)#

#= 1/2(ln8-ln5) = 1/2ln(8/5)#

Aug 16, 2017

Option (E) None of the above.

Explanation:

Let us rewrite the given integral as, #I=int_1^2dx/(x(x^2+1)).#

We solve it without Trig. Substn.

We subst. #x^2=t," giving, "2xdx=dt.#

Further, #x=1 rArr t=1, and, x=2 rArr t=4.#

# :. I=1/2int_1^2(2xdx)/{x^2(x^2+1),#

#=1/2int_1^4dt/{t(t+1)}=1/2int_1^4{(t+1)-t}/{t(t+1)}dt,#

#=1/2int_1^4{(t+1)/(t(t+1))-t/(t(t+1))}dt,#

#=1/2int_1^4{1/t-1/(t+1)}dt,#

#=1/2[ln|t|-ln|t+1|]_1^4,#

#=1/2[ln|t/(t+1)|]_1^4,#

#=1/2[ln(4/5)-ln(1/2)],#

#=1/2ln(4/5-:1/2)=1/2ln(8/5),#

#=1/2ln8-1/2ln5,#

#=1/2ln(2^3)-1/2ln5,#

# rArr I=3/2ln2-1/2ln5.#

Clearly, the Right Choice is Option (E) None of the above.

Enjoy Maths.!

Aug 16, 2017

#intdx/(x^3+x)=intdx/(x(x^2+1))#

Let's try the trigonometric substitution #x=tantheta# so we have the simplification #x^2+1=tan^2theta+1=sec^2theta#. This also implies that #dx=sec^2thetad theta#.

So, plugging these in, the integral becomes:

#=int(sec^2thetad theta)/(tanthetasec^2theta)=intcostheta/sinthetad theta=lnabssintheta+C#

If #x=tantheta#, then there is a right triangle where the side opposite #theta# is #x# and the adjacent side is #1#, so the hypotenuse must be #sqrt(x^2+1)#. Thus, as sine is opposite over hypotenuse, #sintheta=x//sqrt(x^2+1)#.

#=lnabs(x/sqrt(x^2+1))+C#

Then:

#int_1^2dx/(x^3+x)=[lnabs(x/sqrt(x^2+1))]_1^2=ln(2/sqrt5)-ln(1/sqrt2)#

#=ln(2/sqrt5*sqrt2)=ln(sqrt(8/5))=1/2ln(8/5)#

Which equals all the other provided answers given on Socratic and shows that the answer to your question is (E), none of the above.