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

Mar 17, 2018

$\frac{1}{9} \left(\sqrt{3} \pi - \ln 8\right)$ or $0.37355$

Explanation:

${\int}_{0}^{1} \frac{t}{1 + {t}^{3}} \setminus \mathrm{dt}$

Separate the derivative into individual terms using partial fraction decomposition,

$\frac{t}{1 + {t}^{3}}$

Factorise the denominator using sum of cubes formula,

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

Apply partial fraction decomposition,

$\frac{t}{\left(1 + t\right) \left(1 - t + {t}^{2}\right)} = \frac{A}{1 + t} + \frac{B t + C}{1 - t + {t}^{2}}$

Multiply throughout by $\left(1 + t\right) \left(1 - t + {t}^{2}\right)$,

$t = A \left(1 - t + {t}^{2}\right) + \left(B t + C\right) \left(1 + t\right)$

Let $t = - 1$,

$- 1 = A \left(1 - \left(- 1\right) + {\left(- 1\right)}^{2}\right) + \left(B \left(- 1\right) + C\right) \left(1 + \left(- 1\right)\right)$
$\therefore A = - \frac{1}{3}$

Let $t = 0$,

$0 = - \frac{1}{3} \left(1 - \left(0\right) + {\left(0\right)}^{2}\right) + \left(B \left(0\right) + C\right) \left(1 + \left(0\right)\right)$
$\therefore C = \frac{1}{3}$

Let $t = 1$,

$1 = - \frac{1}{3} \left(1 - \left(1\right) + {\left(1\right)}^{2}\right) + \left(B \left(1\right) + \frac{1}{3}\right) \left(1 + \left(1\right)\right)$
$\therefore B = \frac{1}{3}$

Conclude,

$\frac{t}{\left(1 + t\right) \left(1 - t + {t}^{2}\right)} = \frac{\frac{1}{3}}{1 + t} + \frac{\frac{1}{3} t + \frac{1}{3}}{1 - t + {t}^{2}}$

Substitute back to the expression,

${\int}_{0}^{1} \frac{\frac{1}{3}}{1 + t} + \frac{\frac{1}{3} t + \frac{1}{3}}{1 - t + {t}^{2}} \setminus \mathrm{dt}$

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,

$\textcolor{red}{- \frac{1}{3} {\int}_{0}^{1} \frac{1}{1 + t} \setminus \mathrm{dt}}$

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 $u = \frac{2}{\sqrt{3}} \left(t - \frac{1}{2}\right)$,

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 $u = \frac{2}{\sqrt{3}} \left(t - \frac{1}{2}\right)$,

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,

$\frac{1}{3} \ln 1 - \frac{1}{3} \ln 2 + \frac{1}{\sqrt{3}} \arctan \left(\frac{1}{\sqrt{3}}\right) - \frac{1}{\sqrt{3}} \arctan \left(- \frac{1}{\sqrt{3}}\right)$

Simplify,

$\frac{1}{3} \ln 1 - \frac{1}{3} \ln 2 + \frac{\pi}{6 \sqrt{3}} + \frac{\pi}{6 \sqrt{3}}$

Factorise,

$\frac{1}{9} \left(3 \ln 1 - 3 \ln 2 + \frac{\sqrt{3}}{2} \pi + \frac{\sqrt{3}}{2} \pi\right)$

Simplify,

$\frac{1}{9} \left(\sqrt{3} \pi - \ln 8\right)$ or $0.37355$

Mar 17, 2018

The answer is $= 0.37$

Explanation:

First calculate the indefinite integral

Factorise the denominator

$1 + {t}^{3} = \left(1 + t\right) \left(1 - t + {t}^{2}\right)$

Perform the decomposition into partial fractions

$\frac{t}{1 + {t}^{3}} = \frac{t}{\left(1 + t\right) \left(1 - t + {t}^{2}\right)}$

$= \frac{A}{1 + t} + \frac{B t + C}{1 - t + {t}^{2}}$

$= \frac{A \left(1 - t + {t}^{2}\right) + \left(B t + C\right) \left(1 + t\right)}{\left(1 + t\right) \left(1 - t + {t}^{2}\right)}$

The denominators are the same, compare the numerators

$t = A \left(1 - t + {t}^{2}\right) + \left(B t + C\right) \left(1 + t\right)$

Let $t = - 1$, $\implies$, $- 1 = 3 A$, $\implies$, $A = - \frac{1}{3}$

Let $t = 0$, $\implies$, $0 = A + C$, $\implies$, $C = - A = \frac{1}{3}$

Coefficients of ${t}^{2}$

$0 = A + B$, $\implies$, $B = - A = \frac{1}{3}$

Finally,

$\frac{t}{1 + {t}^{3}} = \frac{- \frac{1}{3}}{1 + t} + \frac{\frac{1}{3} t + \frac{1}{3}}{1 - t + {t}^{2}}$

$\int \frac{t \mathrm{dt}}{1 + {t}^{3}} = - \frac{1}{3} \int \frac{\mathrm{dt}}{1 + t} + \frac{1}{3} \int \frac{\left(t + 1\right) \mathrm{dt}}{1 - t + {t}^{2}}$

The first integral is

$- \frac{1}{3} \int \frac{\mathrm{dt}}{1 + t} = - \frac{1}{3} \ln \left(1 + t\right)$

$t + 1 = \frac{1}{2} \left(2 t - 1\right) + \frac{3}{2}$

$\frac{1}{3} \int \frac{\left(t + 1\right) \mathrm{dt}}{1 - t + {t}^{2}} = \frac{1}{6} \int \frac{\left(2 t - 1\right) \mathrm{dt}}{1 - t + {t}^{2}} + \frac{1}{2} \int \frac{\mathrm{dt}}{1 - t + {t}^{2}}$

The second integral is

$\frac{1}{6} \int \frac{\left(2 t - 1\right) \mathrm{dt}}{1 - t + {t}^{2}} = \frac{1}{6} \ln \left(1 - t + {t}^{2}\right)$

$1 - t + {t}^{2} = {t}^{2} - t + 1 = {\left(t - \frac{1}{2}\right)}^{2} + \frac{3}{4} =$

Therefore,

$\frac{1}{2} \int \frac{\mathrm{dt}}{1 - t + {t}^{2}} = \frac{1}{2} \int \frac{\mathrm{dt}}{{\left(t - \frac{1}{2}\right)}^{2} + \frac{3}{4}}$

Let $u = \frac{2 t - 1}{\sqrt{3}}$, $\implies$, $\mathrm{du} = \frac{2}{\sqrt{3}} \mathrm{dt}$

$\frac{1}{2} \int \frac{\mathrm{dt}}{1 - t + {t}^{2}} = \frac{1}{2} \int \frac{2 \sqrt{3} \mathrm{du}}{3 {u}^{2} + 3}$

$= \frac{1}{\sqrt{3}} \int \frac{\mathrm{du}}{{u}^{2} + 1}$

$= \frac{1}{\sqrt{3}} \arctan \left(u\right)$

$= \frac{1}{\sqrt{3}} \arctan \left(\frac{2 t - 1}{\sqrt{3}}\right)$

Putting all together

$\int \frac{t \mathrm{dt}}{1 + {t}^{3}} = - \frac{1}{3} \ln \left(1 + t\right) + \frac{1}{6} \ln \left(| 1 - t + {t}^{2} |\right) + \frac{1}{\sqrt{3}} \arctan \left(\frac{2 t - 1}{\sqrt{3}}\right) + C$

Now calculate the definite integral

${\int}_{0}^{1} \frac{t \mathrm{dt}}{1 + {t}^{3}} = {\left[- \frac{1}{3} \ln \left(1 + t\right) + \frac{1}{6} \ln \left(| 1 - t + {t}^{2} |\right) + \frac{1}{\sqrt{3}} \arctan \left(\frac{2 t - 1}{\sqrt{3}}\right)\right]}_{0}^{1}$

$= \left(- \frac{1}{3} \ln 2 + \frac{1}{6} \ln \left(1\right) + \frac{1}{\sqrt{3}} \arctan \left(\frac{1}{\sqrt{3}}\right)\right) - \left(\frac{1}{3} \ln \left(1\right) + \frac{1}{6} \ln \left(1\right) + \frac{1}{\sqrt{3}} \arctan \left(- \frac{1}{\sqrt{3}}\right)\right)$

$= 0.37$

Mar 17, 2018

$\frac{\pi}{3 \sqrt{3}} - \frac{1}{3} \ln 2$

Explanation:

$I = {\int}_{0}^{1} \frac{t}{{t}^{3} + 1} \mathrm{dt} = {\int}_{0}^{1} \frac{t}{\left(t + 1\right) \left({t}^{2} - t + 1\right)} \mathrm{dt}$
Here, $\frac{t}{\left(t + 1\right) \left({t}^{2} - t + 1\right)} = \frac{A}{t + 1} + \frac{B t + C}{{t}^{2} - t + 1}$
$\implies t = A \left({t}^{2} - t + 1\right) + \left(B t + C\right) \left(t + 1\right)$$\implies t = \left(A + B\right) {t}^{2} + \left(- A + B + C\right) t + \left(A + C\right)$
Comparing coefficient of ${t}^{2} , t \mathmr{and}$ constant term,
$A + B = 0 , - A + B + C = 1 , A + C = 0$$\implies A = - \frac{1}{3} , B = C = \frac{1}{3}$, (solve)
$I = {\int}_{0}^{1} \frac{- \frac{1}{3}}{t + 1} \mathrm{dt} + \frac{1}{3} {\int}_{0}^{1} \frac{t + 1}{{t}^{2} - t + 1} \mathrm{dt}$
$I = - \frac{1}{3} {\left[\ln | t + 1 |\right]}_{0}^{1} + \frac{1}{3} \cdot \frac{1}{2} {\int}_{0}^{1} \frac{2 t + 2}{{t}^{2} - t + 1} \mathrm{dt}$
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
$I = - \frac{1}{3} \ln 2 + \frac{1}{6} {\left[\ln | {t}^{2} - t + 1 |\right]}_{0}^{1} + \frac{1}{2} \cdot \frac{1}{\frac{\sqrt{3}}{2}} {\left[{\tan}^{-} 1 \left(\frac{t - \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)\right]}_{0}^{1}$
$I = - \frac{1}{3} \ln 2 + \frac{1}{6} \left[\ln 1 - \ln 1\right] + \frac{1}{\sqrt{3}} \left[{\tan}^{-} 1 \left(\frac{1}{\sqrt{3}}\right) - {\tan}^{-} 1 \left(- \frac{1}{\sqrt{3}}\right)\right]$
$I = - \frac{1}{3} \ln 2 + 0 + \frac{1}{\sqrt{3}} \left(\frac{\pi}{6} + \frac{\pi}{6}\right) = - \frac{1}{3} \ln 2 + \frac{\pi}{3 \sqrt{3}}$