Question

∫1 / [(x)^m +(x)^n] dx =? show process please.? where m and n are integers

Answer 1

Answer on Question #51422 – Math – integral Calculus

$\int 1 / [(x)^n m + (x)^n n] \, dx = ?$ show process please ($m, n$ are integers).

Solution

Denote constant item by $c$.

1) $m = n$

$$\int \frac {d x}{2 x ^ {n}} = \frac {1}{2} \int \frac {d x}{x ^ {n}} = \left[ \begin{array}{l} \frac {1}{2} \ln | x | + c, n = 1 \\ \frac {x ^ {1 - n}}{2 (1 - n)} + c, n \neq 1 \end{array} \right.$$

2) $m \neq n$

Assume that $m > n$. Denote $k = m - n$.

So:

$$\int \frac {d x}{x ^ {m} + x ^ {n}} = \int \frac {d x}{x ^ {n + k} + x ^ {n}} = \int \frac {d x}{x ^ {n} (x ^ {k} + 1)};$$

It follows from the fundamental theorem of algebra, that we can expand $x^k + 1$ into a product of linear and quadratic polynomials with real coefficients:

$$x ^ {k} + 1 = \left[ \begin{array}{c} (x + 1) (x ^ {2} + a _ {1} x + b _ {1}) \dots (x ^ {2} + a _ {p} x + b _ {p}), 2 \nmid k \\ (x ^ {2} + a _ {1} x + b _ {1}) \dots (x ^ {2} + a _ {p} x + b _ {p}), 2 \mid k \end{array} \right.;$$

So we can expand fraction $\frac{1}{x^n(x^k + 1)}$ into the following sum:

$$\frac {1}{x ^ {n} (x ^ {k} + 1)} = \frac {c _ {1}}{x} + \dots + \frac {c _ {n}}{x ^ {n}} + \frac {\delta (k) c _ {n + 1}}{x + 1} + \frac {d _ {1} x + f _ {1}}{x ^ {2} + a _ {1} x + b _ {1}} + \dots + \frac {d _ {p} x + f _ {p}}{x ^ {2} + a _ {p} x + b _ {p}},$$

where all coefficients $c_{i},d_{i},f_{i}$ are real and \delta (k) = \left[ \begin{array}{l}1,2\notin k\\ 0,2\mid k \end{array} \right;

Hence:

$$\begin{array}{l} \int \frac {d x}{x ^ {n} (x ^ {k} + 1)} = c _ {1} \int \frac {d x}{x} + \dots + c _ {n} \int \frac {d x}{x ^ {n}} + \delta (k) c _ {n + 1} \int \frac {d x}{x + 1} + \int \frac {d _ {1} x + f _ {1}}{x ^ {2} + a _ {1} x + b _ {1}} d x + \\ + \dots + \int \frac {d _ {p} x + f _ {p}}{x ^ {2} + a _ {p} x + b _ {p}} d x; \\ \end{array}$$

The first $n$ integrals are computed in 1).

$$\int \frac {d x}{x + 1} = \ln | x + 1 | + c;$$

The last $p$ integrals can be computed in the following way:

$$\begin{array}{l} \int \frac {d x + f}{x ^ {2} + a x + b} d x = \int \frac {d x + f}{\left(x + \frac {a}{2}\right) ^ {2} + b - \frac {a ^ {2}}{4}} d x = \left[ x + \frac {a}{2} = y \right] = \\ = \int \frac {d ^ {\prime} y + f ^ {\prime}}{y ^ {2} + b - \frac {a ^ {2}}{4}} d y = \frac {d ^ {\prime}}{2} \int \frac {d (y ^ {2})}{y ^ {2} + b - \frac {a ^ {2}}{4}} + f ^ {\prime} \int \frac {d y}{y ^ {2} + b - \frac {a ^ {2}}{4}} = \\ = \frac {d ^ {\prime}}{2} \ln \left| y ^ {2} + b - \frac {a ^ {2}}{4} \right| + c + f ^ {\prime} \int \frac {d y}{y ^ {2} + b - \frac {a ^ {2}}{4}} = \\ \end{array}$$ $$= \frac {d ^ {\prime}}{2} \ln \left| (x + \frac {a}{2}) ^ {2} + b - \frac {a ^ {2}}{4} \right| + c + f ^ {\prime} \int \frac {d y}{y ^ {2} + b - \frac {a ^ {2}}{4}};$$ $$\int \frac {d y}{y ^ {2} + b - \frac {a ^ {2}}{4}} = \left\{ \begin{array}{c} \frac {1}{\sqrt {b - \frac {a ^ {2}}{4}}} \arctan \frac {y}{\sqrt {b - \frac {a ^ {2}}{4}}} + c, b - \frac {a ^ {2}}{4} > 0 \\ - \frac {1}{y} + c, b - \frac {a ^ {2}}{4} = 0 \\ \frac {1}{2 \sqrt {b - \frac {a ^ {2}}{4}}} \ln | \frac {x - \sqrt {b - \frac {a ^ {2}}{4}}}{x + \sqrt {b - \frac {a ^ {2}}{4}}} |, b - \frac {a ^ {2}}{4} < 0 \end{array} \right.$$

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