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

3 Answers
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.

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

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))