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Evaluation of $\displaystyle \int\frac{x^3+3x+2}{(x^2+1)^2\cdot (x+1)}\;,$ Can we solve it without Using Partial fraction.

$\bf{My\; Try::}$ Let $\displaystyle I = \int\frac{x^3+x+2x+2}{(x^2+1)^2\cdot (x+1)}dx = \int\frac{x(x^2+1)+2(x+1)}{(x^2+1)^2(x+1)}dx$

So Integral $\displaystyle I = \int\frac{x}{(x+1)(x^2+1)}dx+2\int\frac{1}{(x^2+1)^2}dx = \int\frac{(x+1)-1}{(x+1)(x^2+1)}dx+2\int\frac{1}{(x^2+1)^2}dx$

So Integral $\displaystyle I = 3\int\frac{1}{(x^2+1)^2}dx-\int\frac{1}{(x+1)(x^2+1)}dx$

Let $\displaystyle I = J-K\;,$ Where $\displaystyle J = 3\int\frac{1}{(x^2+1)^2}dx$ and $\displaystyle J = \int \frac{x}{(x+1)(x^2+1)}$ Now For Calculation of $\displaystyle J = 3\int\frac{1}{(x^2+1)^2}dx\;,$ Let $x = \tan \theta$ and $dx = \sec^2 \theta d\theta$

So Integral

$\displaystyle J = 3\int\frac{\sec^2 \theta}{\sec^4 \theta }d\theta = \frac{3}{2}\int (1+\cos 2\theta )d\theta = \frac{3}{2}\theta + \frac{3}{4}\sin 2\theta = \frac{3}{2}\tan^{-1}\left(x\right)+\frac{3}{2}\frac{x}{1+x^2}$

Now How can I solve $\displaystyle K = \int\frac{1}{(x+1)(x^2+1)}dx = \int\frac{1}{1+\tan \theta}d\theta = \int\frac{\cos \theta}{\sin \theta + \cos \theta }d\theta$

How can I solve after that, Help me

Thanks

juantheron
  • 56,203

1 Answers1

1

You can avoid both partial fraction decomposition and trigonometric substitutions by using the magic rational substitution

$$y=\frac{1-x}{1+x}\iff x=\frac{1-y}{1+y}.$$

Factors of $(x+1)$ become

$$x+1=\left(\frac{1-y}{1+y}\right)+1=\frac{1-y}{1+y}+\frac{1+y}{1+y}=\frac{2}{1+y},$$

and factors of $(x^2+1)$ become

$$x^2+1=\left(\frac{1-y}{1+y}\right)^2+1=\frac{(1-y)^2}{(1+y)^2}+\frac{(1+y)^2}{(1+y)^2}=\frac{2(1+y^2)}{(1+y)^2}.$$

Assuming $x\neq-1$, the integral then becomes under the substitution:

$$\begin{align} I &=\int\frac{x^3+3x+2}{(x^2+1)^2(x+1)}\,\mathrm{d}x\\ &=\int\frac{x(x^2+1)+2(x+1)}{(x^2+1)^2(x+1)}\,\mathrm{d}x\\ &=\int\frac{\left(\frac{1-y}{1+y}\right)\left(\frac{2(1+y^2)}{(1+y)^2}\right)+2\left(\frac{2}{1+y}\right)}{\left(\frac{2(1+y^2)}{(1+y)^2}\right)^2\left(\frac{2}{1+y}\right)}\cdot\frac{(-2)\,\mathrm{d}y}{(1+y)^2}\\ &=\int\frac{(1+y)^2(3+3y+3y^2-y^3)}{4(1+y^2)^2}\cdot\frac{(-2)\,\mathrm{d}y}{(1+y)^2}\\ &=\frac12\int\frac{y^3-3y^2-3y-3}{(1+y^2)^2}\,\mathrm{d}y\\ &=\frac12\int\frac{y(y^2+1)-4y-3(y^2+1)}{(y^2+1)^2}\,\mathrm{d}y\\ &=\frac12\int\frac{y-3}{y^2+1}\,\mathrm{d}y-\int\frac{2y}{(y^2+1)^2}\,\mathrm{d}y\\ &=\frac14\int\frac{2y}{y^2+1}\,\mathrm{d}y-\frac32\int\frac{\mathrm{d}y}{y^2+1}-\int\frac{2y}{(y^2+1)^2}\,\mathrm{d}y\\ &=\frac14\ln{\left(y^2+1\right)}-\frac32\arctan{\left(y\right)}+\frac{1}{y^2+1}+\color{grey}{constant}\\ &=\frac14\ln{\left(\frac{2(x^2+1)}{(x+1)^2}\right)}-\frac32\arctan{\left(\frac{1-x}{1+x}\right)}+\frac{(x+1)^2}{2(x^2+1)}+\color{grey}{constant}.\\ \end{align}$$

David H
  • 32,536