Evaluate #int_0^1 (t/(1+t^3))dt#?

3 Answers
Evan
Mar 17, 2018

#1/9(sqrt3pi-ln8)# or #0.37355#

Explanation:

#int_0^1 t/(1+t^3) \ dt#

Separate the derivative into individual terms using partial fraction decomposition,

#t/(1+t^3)#

Factorise the denominator using sum of cubes formula,

#t/((1+t)(1-t+t^2)#

Apply partial fraction decomposition,

#t/((1+t)(1-t+t^2))=A/(1+t)+(Bt+C)/(1-t+t^2)#

Multiply throughout by #(1+t)(1-t+t^2)#,

#t=A(1-t+t^2)+(Bt+C)(1+t)#

Let #t=-1#,

#-1=A(1-(-1)+(-1)^2)+(B(-1)+C)(1+(-1))#
#:.A=-1/3#

Let #t=0#,

#0=-1/3(1-(0)+(0)^2)+(B(0)+C)(1+(0))#
#:.C=1/3#

Let #t=1#,

#1=-1/3(1-(1)+(1)^2)+(B(1)+1/3)(1+(1))#
#:.B=1/3#

Conclude,

#t/((1+t)(1-t+t^2))=(1/3)/(1+t)+(1/3t+1/3)/(1-t+t^2)#

Substitute back to the expression,

#int_0^1 (1/3)/(1+t)+(1/3t+1/3)/(1-t+t^2) \ dt#

Apply sum rule and take the constants out,

#color(red)(-1/3int_0^1 1/(1+t) \ dt)+color(blue)(1/3int_0^1(t+1)/(1-t+t^2) \ dt#

Let's separate the terms to make them easier to work with,

  • First integral,

#color(red)(-1/3int_0^1 1/(1+t) \ dt)#

Integrate,

#color(red)([-1/3ln|1+t|]_0^1#

  • Second integral,

#color(blue)(1/3int_0^1(t+1)/(1-t+t^2) \ dt#

Complete the square for the denominator,

#color(blue)(1/3int_0^1(t+1)/((t-1/2)^2+3/4) \ dt#

Multiply the function by #2#, by dividing the constant by #2#,

#color(blue)(1/6int_0^1(2t+2)/((t-1/2)^2+3/4) \ dt#

Apply sum rule,

#color(blue)(1/6int_0^1(2t-1)/((t-1/2)^2+3/4) \ dt + 1/6int_0^1(3)/((t-1/2)^2+3/4) \ dt#

Integrate first function,

#color(blue)([1/6ln|1-t+t^2|]_0^1 + 1/2int_0^1(1)/((t-1/2)^2+3/4) \ dt#

Apply #u#-substitution, where #u=2/sqrt3(t-1/2)#,

#color(blue)([1/6ln|1-t+t^2|]_0^1 + 1/2int_0^1(sqrt3/2)/(3/4u^2+3/4) \ du#

Simplify,

#color(blue)([1/6ln|1-t+t^2|]_0^1 + 1/2int_0^1(2sqrt3)/(3(u^2+1)) \ du#

Take the constant out,

#color(blue)([1/6ln|1-t+t^2|]_0^1 + 1/sqrt3int_0^1(1)/(u^2+1) \ du#

Apply arctangent rule,

#color(blue)([1/6ln|1-t+t^2|]_0^1 + [1/sqrt3arctanu]_0^1#

Substitute back #u=2/sqrt3(t-1/2)#,

#color(blue)([1/6ln|1-t+t^2|]_0^1 + [1/sqrt3arctan((2t-1)/sqrt3)]_0^1#

Substitute the two integrals back,

#color(red)([-1/3ln|1+t|]_0^1)+color(blue)([1/6ln|1-t+t^2|]_0^1 + [1/sqrt3arctan((2t-1)/sqrt3)]_0^1#

Expand,

#color(red)((-1/3ln|2|)-(-1/3ln|1|))+color(blue)((1/6ln|1|)-(1/6ln|1|) + (1/sqrt3arctan(1/sqrt3))-(1/sqrt3arctan(-1/sqrt3))#

Remove parenthesis,

#1/3ln1-1/3ln2+ 1/sqrt3arctan(1/sqrt3)-1/sqrt3arctan(-1/sqrt3)#

Simplify,

#1/3ln1-1/3ln2+ pi/(6sqrt3)+pi/(6sqrt3)#

Factorise,

#1/9(3ln1-3ln2+ sqrt3/2pi+sqrt3/2pi)#

Simplify,

#1/9(sqrt3pi-ln8)# or #0.37355#

Tad bit long.

Narad T.
Mar 17, 2018

The answer is #=0.37#

Explanation:

First calculate the indefinite integral

Factorise the denominator

#1+t^3=(1+t)(1-t+t^2)#

Perform the decomposition into partial fractions

#t/(1+t^3)=(t)/((1+t)(1-t+t^2))#

#=A/(1+t)+(Bt+C)/(1-t+t^2)#

#=(A(1-t+t^2)+(Bt+C)(1+t))/((1+t)(1-t+t^2))#

The denominators are the same, compare the numerators

#t=A(1-t+t^2)+(Bt+C)(1+t)#

Let #t=-1#, #=>#, #-1=3A#, #=>#, #A=-1/3#

Let #t=0#, #=>#, #0=A+C#, #=>#, #C=-A=1/3#

Coefficients of #t^2#

#0=A+B#, #=>#, #B=-A=1/3#

Finally,

#t/(1+t^3)=(-1/3)/(1+t)+(1/3t+1/3)/(1-t+t^2)#

#int(tdt)/(1+t^3)=-1/3int(dt)/(1+t)+1/3int((t+1)dt)/(1-t+t^2)#

The first integral is

#-1/3int(dt)/(1+t)=-1/3ln(1+t)#

#t+1=1/2(2t-1)+3/2#

#1/3int((t+1)dt)/(1-t+t^2)=1/6int((2t-1)dt)/(1-t+t^2)+1/2int(dt)/(1-t+t^2)#

The second integral is

#1/6int((2t-1)dt)/(1-t+t^2)=1/6ln(1-t+t^2)#

#1-t+t^2=t^2-t+1= (t-1/2)^2+3/4=#

Therefore,

#1/2int(dt)/(1-t+t^2)=1/2int(dt)/((t-1/2)^2+3/4)#

Let #u=(2t-1)/sqrt3#, #=>#, #du=2/sqrt3dt#

#1/2int(dt)/(1-t+t^2)=1/2int(2sqrt3du)/(3u^2+3)#

#=1/sqrt3int(du)/(u^2+1)#

#=1/sqrt3arctan(u)#

#=1/sqrt3arctan((2t-1)/sqrt3)#

Putting all together

#int(tdt)/(1+t^3)=-1/3ln(1+t)+1/6ln(|1-t+t^2|)+1/sqrt3arctan((2t-1)/sqrt3)+C#

Now calculate the definite integral

#int_0^1(tdt)/(1+t^3)=[-1/3ln(1+t)+1/6ln(|1-t+t^2|)+1/sqrt3arctan((2t-1)/sqrt3)]_0^1#

#=(-1/3ln2+1/6ln(1)+1/sqrt3arctan(1/sqrt3))-(1/3ln(1)+1/6ln(1)+1/sqrt3arctan(-1/sqrt3))#

#=0.37#

maganbhai P.
Mar 17, 2018

#pi/(3sqrt(3))-1/3ln2#

Explanation:

#I=int_0^1t/(t^3+1)dt=int_0^1t/((t+1)(t^2-t+1))dt#
Here, #t/((t+1)(t^2-t+1))=A/(t+1)+(Bt+C)/(t^2-t+1)#
#=>t=A(t^2-t+1)+(Bt+C)(t+1)##=>t=(A+B)t^2+(-A+B+C)t+(A+C)#
Comparing coefficient of #t^2,t and# constant term,
#A+B=0,-A+B+C=1,A+C=0##=>A=-1/3,B=C=1/3#, (solve)
#I=int_0^1(-1/3)/(t+1)dt+1/3int_0^1(t+1)/(t^2-t+1)dt#
#I=-1/3[ln|t+1|]_0^1+1/3*1/2int_0^1(2t+2)/(t^2-t+1)dt#
#I=-1/3[ln2-ln1]+1/6int_0^1(2t-1)/(t^2-t+1)dt+1/6int_0^1 3/((t-1/2)^2+(sqrt(3)/2)^2#
#I=-1/3ln2+1/6[ln|t^2-t+1|]_0^1+1/2*1/(sqrt(3)/2)[tan^-1((t-1/2)/(sqrt(3)/2))]_0^1#
#I=-1/3ln2+1/6[ln1-ln1]+1/sqrt(3)[tan^-1(1/sqrt(3))-tan^-1(-1/sqrt(3))]#
#I=-1/3ln2+0+1/sqrt(3)(pi/6+pi/6)=-1/3ln2+pi/(3sqrt(3))#

Related questions
Impact of this question
2849 views around the world
You can reuse this answer
Creative Commons License