Evaluate the integral. {eq}\displaystyle \int_0^1 \dfrac{ dx}{(x + 1) (x^2+ 1) } {/eq}

Question:

Evaluate the integral.

{eq}\displaystyle \int_0^1 \dfrac{ dx}{(x + 1) (x^2+ 1) } {/eq}

Partial Fraction:

While obtaining the partial fraction of any rational function, it is important to know whether the degree of the numerator is less than the denominator or not. If not, then we will divide the numerator by the denominator. Many methods are used to integrate functions. These methods include the substitution method, partial fractions method, and integration by parts method.

For the given integral, in a fraction {eq}\deg \left( Nr \right)<\deg \left( Dr \right) {/eq}.

Thus, apply partial fraction decomposition to the fraction {eq}\dfrac{1}{\left(x+1\right)\left(x^2+1\right)} {/eq}.

{eq}\dfrac{1}{\left(x+1\right)\left(x^2+1\right)}=\dfrac{a_0}{x+1}+\dfrac{a_2x+a_1}{x^2+1} {/eq}

Simplify the left side numerator of the equation {eq}\dfrac{1}{\left(x+1\right)\left(x^2+1\right)}=\dfrac{a_0}{x+1}+\dfrac{a_2x+a_1}{x^2+1} {/eq} in {eq}ax^2+bx+c {/eq} form.

{eq}\begin{aligned} \dfrac{1}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}&=\dfrac{{{a}_{0}}}{x+1}+\dfrac{{{a}_{2}}x+{{a}_{1}}}{{{x}^{2}}+1} \\ &=\dfrac{{{a}_{0}}\left( {{x}^{2}}+1 \right)+\left( {{a}_{2}}x+{{a}_{1}} \right)\left( x+1 \right)}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)} \\ &=\dfrac{{{a}_{0}}{{x}^{2}}+{{a}_{0}}+{{a}_{2}}{{x}^{2}}+{{a}_{1}}x+{{a}_{2}}x+{{a}_{1}}}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)} \\ &=\dfrac{\left( {{a}_{0}}+{{a}_{2}} \right){{x}^{2}}+\left( {{a}_{1}}+{{a}_{2}} \right)x+{{a}_{0}}+{{a}_{1}}}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)} \\ \end{aligned} {/eq}

Find out the value of {eq}{{a}_{0}} {/eq}, {eq}{{a}_{1}} {/eq}, and {eq}{{a}_{2}} {/eq} by comparing the numerator of the obtained equation.

{eq}\begin{aligned} {{a}_{0}}+{{a}_{2}}&=0 \\ {{a}_{0}}&=-{{a}_{2}} \\ \end{aligned} {/eq}

{eq}\begin{aligned} {{a}_{1}}+{{a}_{2}}&=0 \\ {{a}_{1}}&=-{{a}_{2}} \\ \end{aligned} {/eq}

{eq}\begin{aligned} {{a}_{0}}+{{a}_{1}}&=1 \\ -{{a}_{2}}-{{a}_{2}}&=1\,\,\,\,\,\,\,\,\,\,\left\{ \because {{a}_{0}}=-{{a}_{2}},{{a}_{1}}=-{{a}_{2}} \right\} \\ -2{{a}_{2}}&=1 \\ {{a}_{2}}&=-\dfrac{1}{2} \end{aligned} {/eq}

It is obtained that {eq}{{a}_{0}}=-{{a}_{2}} {/eq}, {eq}{{a}_{1}}=-{{a}_{2}} {/eq}, and {eq}{{a}_{2}}=-\dfrac{1}{2} {/eq}. So, {eq}{{a}_{0}}=\dfrac{1}{2} {/eq} and {eq}{{a}_{1}}=\dfrac{1}{2} {/eq}.

Substitute the obtained value of {eq}{{a}_{0}} {/eq}, {eq}{{a}_{1}} {/eq}, and {eq}{{a}_{2}} {/eq} into the equation {eq}\dfrac{1}{\left(x+1\right)\left(x^2+1\right)}=\dfrac{a_0}{x+1}+\dfrac{a_2x+a_1}{x^2+1} {/eq} and simplify.

{eq}\begin{aligned} \dfrac{1}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}&=\dfrac{\left( \dfrac{1}{2} \right)}{x+1}+\dfrac{\left( -\dfrac{1}{2} \right)x+\left( \dfrac{1}{2} \right)}{{{x}^{2}}+1} \\ &=\dfrac{1}{2\left( x+1 \right)}+\dfrac{-x+1}{2\left( {{x}^{2}}+1 \right)} \end{aligned} {/eq}

Substitute {eq}\dfrac{1}{2\left( x+1 \right)}+\dfrac{-x+1}{2\left( {{x}^{2}}+1 \right)} {/eq} for {eq}\dfrac{1}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)} {/eq} into the given integral and simplify.

{eq}\begin{aligned} \displaystyle\int_{0}^{1}{\dfrac{1}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}dx}&=\displaystyle\int_{0}^{1}{\left( \dfrac{1}{2\left( x+1 \right)}+\dfrac{-x+1}{2\left( {{x}^{2}}+1 \right)} \right)dx} \\ &=\displaystyle\int_{0}^{1}{\left( \dfrac{1}{2\left( x+1 \right)} \right)dx}+\displaystyle\int_{0}^{1}{\left( \dfrac{-x+1}{2\left( {{x}^{2}}+1 \right)} \right)dx} \\ &=\dfrac{1}{2}\displaystyle\int_{0}^{1}{\left( \dfrac{1}{x+1} \right)dx}+\dfrac{1}{2}\displaystyle\int_{0}^{1}{\left( \dfrac{-x}{{{x}^{2}}+1} \right)dx}+\dfrac{1}{2}\displaystyle\int_{0}^{1}{\left( \dfrac{1}{{{x}^{2}}+1} \right)dx} \\ &=\dfrac{1}{2}\left[ \ln \left( x+1 \right) \right]_{0}^{1}-\dfrac{1}{2}\displaystyle\int_{0}^{1}{\left( \dfrac{x}{{{x}^{2}}+1} \right)dx}+\dfrac{1}{2}\left[ \arctan \left( x \right) \right]_{0}^{1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \because \displaystyle\int{\dfrac{1}{{{x}^{2}}+1}dx=}\arctan \left( x \right) \right\} \\ &=\dfrac{1}{2}\left[ \ln \left( 1+1 \right)-\ln \left( 0+1 \right) \right]-\dfrac{1}{2}\displaystyle\int_{1}^{2}{\dfrac{1}{2u}du}+\dfrac{1}{2}\left[ \arctan \left( 1 \right)-\arctan \left( 0 \right) \right]\,\,\,\,\,\,\,\left\{ \text{Let}\,\,u={{x}^{2}}+1\,\,\text{then}\,\,du=2x \right\} \\ &=\dfrac{1}{2}\left[ \ln \left( 2 \right)-\ln \left( 1 \right) \right]-\dfrac{1}{4}\displaystyle\int_{1}^{2}{\dfrac{1}{u}du}+\dfrac{1}{2}\left[ \dfrac{\pi }{4}-0 \right] \\ &=\dfrac{1}{2}\left[ \ln \left( 2 \right)-0 \right]-\dfrac{1}{4}\left[ \ln \left( u \right) \right]_{1}^{2}+\dfrac{\pi }{8} \\ &=\dfrac{1}{2}\ln \left( 2 \right)-\dfrac{1}{4}\left[ \ln \left( 2 \right)-\ln \left( 1 \right) \right]+\dfrac{\pi }{8} \\ &=\dfrac{1}{2}\ln \left( 2 \right)-\dfrac{1}{4}\left[ \ln \left( 2 \right)-0 \right]+\dfrac{\pi }{8} \\ &=\dfrac{1}{2}\ln \left( 2 \right)-\dfrac{1}{4}\ln \left( 2 \right)+\dfrac{\pi }{8} \\ &=\dfrac{4\ln \left( 2 \right)-2\ln \left( 2 \right)+\pi }{8} \\ &=\dfrac{2\ln \left( 2 \right)+\pi }{8} \end{aligned} {/eq}

Thus, {eq}\displaystyle\int_{0}^{1}{\dfrac{1}{\left( x+1 \right)\left( {{x}^{2}}+1 \right)}dx}=\dfrac{\mathbf{2ln}\left( \mathbf{2} \right)\mathbf{+\pi }}{\mathbf{8}} {/eq}.