Understanding the Integral Problem

The question asks us to find the indefinite integral of the function \( \frac{1}{{{\left( {x + 1} \right)}^2}\left( {{x^2} + 1} \right)}} \) with respect to \(x\). This type of integral, involving a rational function where the denominator is a product of linear and irreducible quadratic factors, is typically solved using the method of partial fraction decomposition.

Partial Fraction Decomposition Method

The first step in solving this integral is to decompose the integrand into simpler fractions. The denominator has a repeated linear factor \( (x+1)^2 \) and an irreducible quadratic factor \( (x^2+1) \). Therefore, the partial fraction form will be:

$$ \frac{1}{{{\left( {x + 1} \right)}^2}\left( {{x^2} + 1} \right)}} = \frac{A}{{x + 1}} + \frac{B}{{{{\left( {x + 1} \right)}^2}}} + \frac{{Cx + D}}{{{x^2} + 1}} $$

To find the values of the constants \(A\), \(B\), \(C\), and \(D\), we clear the denominators by multiplying both sides by \( {{\left( {x + 1} \right)}^2}\left( {{x^2} + 1} \right)} \):

$$ 1 = A(x + 1)({x^2} + 1) + B({x^2} + 1) + (Cx + D){(x + 1)^2} $$

Solving for Coefficients (A, B, C, D)

We can find the constants by substituting specific values for \(x\) or by comparing coefficients of powers of \(x\).

Step 1: Find B by substituting \(x = -1\)

Let \(x = -1\) in the equation:

$$ 1 = A(-1 + 1)((-1)^2 + 1) + B((-1)^2 + 1) + (C(-1) + D)(-1 + 1)^2 $$ $$ 1 = A(0)(2) + B(1 + 1) + (-C + D)(0)^2 $$ $$ 1 = 0 + 2B + 0 $$ $$ 1 = 2B \implies B = \frac{1}{2} $$

Step 2: Expand and Compare Coefficients

Now, let's expand the right side of the equation \( 1 = A(x + 1)({x^2} + 1) + B({x^2} + 1) + (Cx + D){(x + 1)^2} \):

$$ 1 = A(x^3 + x^2 + x + 1) + B(x^2 + 1) + (Cx + D)(x^2 + 2x + 1) $$ $$ 1 = Ax^3 + Ax^2 + Ax + A + Bx^2 + B + Cx^3 + 2Cx^2 + Cx + Dx^2 + 2Dx + D $$

Group terms by powers of \(x\):

$$ 1 = (A + C)x^3 + (A + B + 2C + D)x^2 + (A + C + 2D)x + (A + B + D) $$

Now, we compare the coefficients of the powers of \(x\) on both sides of the equation. Since the left side is just \(1\), the coefficients of \(x^3\), \(x^2\), and \(x\) are zero, and the constant term is one.

• Coefficient of \(x^3\): \(A + C = 0 \quad \text{(Equation 1)}\)
• Coefficient of \(x^2\): \(A + B + 2C + D = 0 \quad \text{(Equation 2)}\)
• Coefficient of \(x^1\): \(A + C + 2D = 0 \quad \text{(Equation 3)}\)
• Constant term (\(x^0\)): \(A + B + D = 1 \quad \text{(Equation 4)}\)

We already found \( B = \frac{1}{2} \).

From Equation 1, \( C = -A \).

Substitute \( C = -A \) into Equation 3:

$$ A + (-A) + 2D = 0 $$ $$ 0 + 2D = 0 \implies D = 0 $$

Now substitute \( B = \frac{1}{2} \) and \( D = 0 \) into Equation 4:

$$ A + \frac{1}{2} + 0 = 1 $$ $$ A = 1 - \frac{1}{2} \implies A = \frac{1}{2} $$

Finally, use \( C = -A \):

$$ C = -\frac{1}{2} $$

So, the partial fraction decomposition is:

$$ \frac{1}{{{\left( {x + 1} \right)}^2}\left( {{x^2} + 1} \right)}} = \frac{{\frac{1}{2}}}{{x + 1}} + \frac{{\frac{1}{2}}}{{{{\left( {x + 1} \right)}^2}}} + \frac{{ - \frac{1}{2}x + 0}}{{{x^2} + 1}} $$ $$ = \frac{1}{{2(x + 1)}} + \frac{1}{{2{{(x + 1)}^2}}} - \frac{x}{{2({x^2} + 1)}} $$

Integrating Each Term

Now we integrate each term separately:

1. Integral of the first term: $$ \smallint \frac{1}{{2(x + 1)}}dx = \frac{1}{2}\smallint \frac{1}{{x + 1}}dx = \frac{1}{2}\log |x + 1| $$
2. Integral of the second term: $$ \smallint \frac{1}{{2{{(x + 1)}^2}}}dx = \frac{1}{2}\smallint {{(x + 1)}^{ - 2}}dx = \frac{1}{2} \left( \frac{{{{(x + 1)}^{ - 1}}}}{{ - 1}} \right) = - \frac{1}{{2(x + 1)}} $$
3. Integral of the third term: $$ \smallint - \frac{x}{{2({x^2} + 1)}}dx = - \frac{1}{2}\smallint \frac{x}{{{x^2} + 1}}dx $$ To solve this, let \( u = x^2 + 1 \). Then \( du = 2x \,dx \), which means \( x \,dx = \frac{1}{2}du \). $$ - \frac{1}{2}\smallint \frac{1}{u} \left( \frac{1}{2} \right)du = - \frac{1}{4}\smallint \frac{1}{u}du = - \frac{1}{4}\log |u| = - \frac{1}{4}\log (x^2 + 1) $$ (Note: \(x^2 + 1\) is always positive, so the absolute value is not strictly necessary here).

Final Solution

Combining all the integrated terms, we get the final indefinite integral:

$$ \smallint \frac{{dx}}{{{{\left( {x + 1} \right)}^2}\left( {{x^2} + 1} \right)}} = \frac{1}{2}\log |x + 1| - \frac{1}{{2(x + 1)}} - \frac{1}{4}\log (x^2 + 1) + C $$

Comparing this result with the given options, the correct one matches:

$$ \frac{1}{2}\log \left( {x + 1} \right) - \frac{1}{{2\left( {x + 1} \right)}} - \frac{1}{4}\log \left( {{x^2} + 1} \right) $$