Question Statement

Evaluate the integral: $\int \frac{x}{(x+1{)}^2\left(x^2+1\right)}dx$

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Background and Explanation

These problems involve integrating rational functions using partial fraction decomposition. For Q11, we first use substitution to simplify the integrand into a rational function in a new variable. For Q12, we directly decompose the rational function into simpler fractions that can be integrated using standard formulas.

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Solution

$\int \frac{x}{(x+1{)}^2\left(x^2+1\right)}dx$

Step 1: Partial Fraction Setup Consider: $\frac{x}{(x+1{)}^2\left(x^2+1\right)}=\frac{A}{x+1}+\frac{B}{(x+1{)}^2}+\frac{Cx+D}{x^2+1}$

Multiplying both sides by $(x+1{)}^2\left(x^2+1\right)$: $x=A(x+1)\left(x^2+1\right)+B\left(x^2+1\right)+(Cx+D)(x+1{)}^2$

Step 2: Finding B Put $x+1=0$ (i.e., $x=-1$) in equation (2): $-1=A(0)(2)+B(2)+(-C+D)(0)$ $-1=2B$ $\frac{-1}{2}=B$

Step 3: Expanding and Comparing Coefficients From equation (2): $x=A(x^3+x^2+x+1)+B(x^2+1)+(Cx+D)(x^2+2x+1)$

Expanding the right side: $x=A{x}^3+A{x}^2+Ax+A+B{x}^2+B+C{x}^3+2C{x}^2+Cx+D{x}^2+2Dx+D$

Grouping by powers: $x=(A+C)x^3+(A+B+2C+D)x^2+(A+C+2D)x+(A+B+D)$

Comparing coefficients of $x^3$: $0=A+C$

Comparing coefficients of $x^2$: $0=A+B+2C+D$

Comparing coefficients of $x$: $1=A+C+2D$

Using equation (3) ($A+C=0$) in the $x$ coefficient equation: $1=0+2D$ $\frac{1}{2}=D$

Step 4: Finding A and C Putting values of $B=-\frac{1}{2}$ and $D=\frac{1}{2}$ in the $x^2$ coefficient equation: $0=A-\frac{1}{2}+2C+\frac{1}{2}$ $0=A+2C$

Subtracting equation (3) from equation (5): $\begin{array}{r}0=A+2C\\ 0=A+C\\ 0=C\end{array}$

Putting $C=0$ in equation (3): $0=A+0$ $0=A$

Step 5: Writing the Decomposition Putting values of $A=0$, $B=-\frac{1}{2}$, $C=0$, and $D=\frac{1}{2}$ in equation (1): $\frac{x}{(x+1{)}^2\left(x^2+1\right)}=\frac{0}{x+1}+\frac{-\frac{1}{2}}{(x+1{)}^2}+\frac{0\cdot x+\frac{1}{2}}{x^2+1}$

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

Step 6: Integration Integrating both sides with respect to $x$: $\int \frac{x}{(x+1{)}^2\left(x^2+1\right)}dx=\frac{-1}{2}\int \frac{1}{(x+1{)}^2}dx+\frac{1}{2}\int \frac{1}{x^2+1}dx$

Evaluating: $=\frac{-1}{2}\int (x+1{)}^{-2}dx+\frac{1}{2}{\tan }^{-1}(x)$ $=\frac{-1}{2}\cdot \frac{(x+1{)}^{-2+1}}{-2+1}+\frac{1}{2}{\tan }^{-1}(x)+C$ $=\frac{-1}{2}\cdot \frac{(x+1{)}^{-1}}{-1}+\frac{1}{2}{\tan }^{-1}(x)+C$ $=\frac{1}{2(x+1)}+\frac{1}{2}{\tan }^{-1}x+C$

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Key Formulas or Methods Used

• Substitution Rule: $\int f(g(x))g^{\prime }(x)dx=\int f(t)dt$ where $t=g(x)$
• Partial Fraction Decomposition:
  • For repeated linear factors: $\frac{P(x)}{(x-a{)}^2(x-b)}=\frac{A}{x-a}+\frac{B}{(x-a{)}^2}+\frac{C}{x-b}$
  • For irreducible quadratic factors: $\frac{P(x)}{(x-a{)}^2(x^2+b^2)}=\frac{A}{x-a}+\frac{B}{(x-a{)}^2}+\frac{Cx+D}{x^2+b^2}$
• Integration Formulas:
  • $\int \frac{1}{x}dx=\ln |x|+C$
  • $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$ (for $n\ne -1$)
  • $\int \frac{1}{x^2+1}dx={\tan }^{-1}x+C$

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Summary of Steps

1. Set up partial fractions: $\frac{A}{x+1}+\frac{B}{(x+1{)}^2}+\frac{Cx+D}{x^2+1}$
2. Find $B=-\frac{1}{2}$ by putting $x=-1$
3. Expand and compare coefficients of $x^3$, $x^2$, and $x$ to find $A=0$, $C=0$, $D=\frac{1}{2}$
4. Rewrite the integrand as $\frac{-1}{2(x+1{)}^2}+\frac{1}{2(x^2+1)}$
5. Integrate using power rule and arctangent formula