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Intégrale de 11−x2\frac{1}{1 - x^{2}}

La calculatrice trouvera l'intégrale/antidérivée de 11−x2\frac{1}{1 - x^{2}}, avec les étapes indiquées.

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Votre contribution

Trouvez ∫11−x2 dx\int \frac{1}{1 - x^{2}}\, dx.

Solution

Perform partial fraction decomposition (steps can be seen »):

∫11−x2dx=∫(12(x+1)−12(x−1))dx{\color{red}{\int{\frac{1}{1 - x^{2}} d x}}} = {\color{red}{\int{\left(\frac{1}{2 \left(x + 1\right)} - \frac{1}{2 \left(x - 1\right)}\right)d x}}}

Integrate term by term:

∫(12(x+1)−12(x−1))dx=(−∫12(x−1)dx+∫12(x+1)dx){\color{red}{\int{\left(\frac{1}{2 \left(x + 1\right)} - \frac{1}{2 \left(x - 1\right)}\right)d x}}} = {\color{red}{\left(- \int{\frac{1}{2 \left(x - 1\right)} d x} + \int{\frac{1}{2 \left(x + 1\right)} d x}\right)}}

Apply the constant multiple rule ∫cf(x) dx=c∫f(x) dx\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx with c=12c=\frac{1}{2} and f(x)=1x+1f{\left(x \right)} = \frac{1}{x + 1}:

−∫12(x−1)dx+∫12(x+1)dx=−∫12(x−1)dx+(∫1x+1dx2)- \int{\frac{1}{2 \left(x - 1\right)} d x} + {\color{red}{\int{\frac{1}{2 \left(x + 1\right)} d x}}} = - \int{\frac{1}{2 \left(x - 1\right)} d x} + {\color{red}{\left(\frac{\int{\frac{1}{x + 1} d x}}{2}\right)}}

Let u=x+1u=x + 1.

Then du=(x+1)′dx=1dxdu=\left(x + 1\right)^{\prime }dx = 1 dx (steps can be seen »), and we have that dx=dudx = du.

C'est pourquoi,

−∫12(x−1)dx+∫1x+1dx2=−∫12(x−1)dx+∫1udu2- \int{\frac{1}{2 \left(x - 1\right)} d x} + \frac{{\color{red}{\int{\frac{1}{x + 1} d x}}}}{2} = - \int{\frac{1}{2 \left(x - 1\right)} d x} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2}

The integral of 1u\frac{1}{u} is ∫1udu=ln⁡(∣u∣)\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}:

−∫12(x−1)dx+∫1udu2=−∫12(x−1)dx+ln⁡(∣u∣)2- \int{\frac{1}{2 \left(x - 1\right)} d x} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2} = - \int{\frac{1}{2 \left(x - 1\right)} d x} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}

Recall that u=x+1u=x + 1:

ln⁡(∣u∣)2−∫12(x−1)dx=ln⁡(∣(x+1)∣)2−∫12(x−1)dx\frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} - \int{\frac{1}{2 \left(x - 1\right)} d x} = \frac{\ln{\left(\left|{{\color{red}{\left(x + 1\right)}}}\right| \right)}}{2} - \int{\frac{1}{2 \left(x - 1\right)} d x}

Apply the constant multiple rule ∫cf(x) dx=c∫f(x) dx\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx with c=12c=\frac{1}{2} and f(x)=1x−1f{\left(x \right)} = \frac{1}{x - 1}:

ln⁡(∣x+1∣)2−∫12(x−1)dx=ln⁡(∣x+1∣)2−(∫1x−1dx2)\frac{\ln{\left(\left|{x + 1}\right| \right)}}{2} - {\color{red}{\int{\frac{1}{2 \left(x - 1\right)} d x}}} = \frac{\ln{\left(\left|{x + 1}\right| \right)}}{2} - {\color{red}{\left(\frac{\int{\frac{1}{x - 1} d x}}{2}\right)}}

Let u=x−1u=x - 1.

Then du=(x−1)′dx=1dxdu=\left(x - 1\right)^{\prime }dx = 1 dx (steps can be seen »), and we have that dx=dudx = du.

The integral can be rewritten as

ln⁡(∣x+1∣)2−∫1x−1dx2=ln⁡(∣x+1∣)2−∫1udu2\frac{\ln{\left(\left|{x + 1}\right| \right)}}{2} - \frac{{\color{red}{\int{\frac{1}{x - 1} d x}}}}{2} = \frac{\ln{\left(\left|{x + 1}\right| \right)}}{2} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2}

The integral of 1u\frac{1}{u} is ∫1udu=ln⁡(∣u∣)\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}:

ln⁡(∣x+1∣)2−∫1udu2=ln⁡(∣x+1∣)2−ln⁡(∣u∣)2\frac{\ln{\left(\left|{x + 1}\right| \right)}}{2} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2} = \frac{\ln{\left(\left|{x + 1}\right| \right)}}{2} - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}

Recall that u=x−1u=x - 1:

ln⁡(∣x+1∣)2−ln⁡(∣u∣)2=ln⁡(∣x+1∣)2−ln⁡(∣(x−1)∣)2\frac{\ln{\left(\left|{x + 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} = \frac{\ln{\left(\left|{x + 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{{\color{red}{\left(x - 1\right)}}}\right| \right)}}{2}

C'est pourquoi,

∫11−x2dx=−ln⁡(∣x−1∣)2+ln⁡(∣x+1∣)2\int{\frac{1}{1 - x^{2}} d x} = - \frac{\ln{\left(\left|{x - 1}\right| \right)}}{2} + \frac{\ln{\left(\left|{x + 1}\right| \right)}}{2}

Simplifier :

∫11−x2dx=−ln⁡(∣x−1∣)+ln⁡(∣x+1∣)2\int{\frac{1}{1 - x^{2}} d x} = \frac{- \ln{\left(\left|{x - 1}\right| \right)} + \ln{\left(\left|{x + 1}\right| \right)}}{2}

Ajouter la constante d'intégration :

∫11−x2dx=−ln⁡(∣x−1∣)+ln⁡(∣x+1∣)2+C\int{\frac{1}{1 - x^{2}} d x} = \frac{- \ln{\left(\left|{x - 1}\right| \right)} + \ln{\left(\left|{x + 1}\right| \right)}}{2}+C

Answer: ∫11−x2dx=−ln⁡(∣x−1∣)+ln⁡(∣x+1∣)2+C\int{\frac{1}{1 - x^{2}} d x}=\frac{- \ln{\left(\left|{x - 1}\right| \right)} + \ln{\left(\left|{x + 1}\right| \right)}}{2}+C

Notes annexes: Substitution (Change of Variable) Rule , Integration by Parts , Concept of Antiderivative and Indefinite Integral , Integrals Involving Trig Functions , Trigonometric Substitutions In Integrals , Integrals Involving Rational Functions , Integration Formulas (Table of Indefinite Integrals) , Properties of Indefinite Integrals , Table of Antiderivatives