int(7)/((x-2)(x+5))dx

Partial Fraction Decomposition: We have the integral to solve: int(7)/((x-2)(x+5))dx To solve this integral, we will use partial fraction decomposition to break the integrand into simpler fractions that can be integrated easily.Express Integrands as Fractions: Express the integrand as a sum of partial fractions. We want to find constants A and B such that: 7/((x-2)(x+5)) = A/(x-2) + B/(x+5) To find A and B, we multiply both sides by the denominator (x-2)(x+5) to get: 7 = A(x+5) + B(x-2)Solve for Constants A and B: Solve for A and B by substituting suitable values for x. Let x = 2, then we get: 7 = A(2+5) + B(2-2) 7 = 7A A = 1 Let x = -5, then we get: 7 = A(-5+5) + B(-5-2) 7 = -7B B = -1Rewrite Integral with Partial Fractions: Rewrite the integral with the found partial fractions. int(7)/((x-2)(x+5))dx = int(1)/(x-2)dx - int(1)/(x+5)dxIntegrate Each Term Separately: Integrate each term separately. The integral of 1/(x-2)dx is ln|x-2|, and the integral of 1/(x+5)dx is ln|x+5|. So, int(1)/(x-2)dx - int(1)/(x+5)dx = ln|x-2| - ln|x+5| + C, where C is the constant of integration.Combine Results for Final Answer: Combine the results to write the final answer. The indefinite integral of 7/((x-2)(x+5)) with respect to x is ln|x-2| - ln|x+5| + C.

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\int \frac{7}{(x-2)(x+5)} d x

Partial Fraction Decomposition: We have the integral to solve: $\newline$ $\int\frac{7}{(x-2)(x+5)}dx$ $\newline$ To solve this integral, we will use partial fractions" target="_blank" class="backlink">fraction decomposition to break the integrand into simpler fractions that can be integrated easily.
Express Integrands as Fractions: Express the integrand as a sum of partial fractions. $\newline$ We want to find constants $A$ and $B$ such that: $\newline$ $\frac{7}{(x-2)(x+5)} = \frac{A}{(x-2)} + \frac{B}{(x+5)}$ $\newline$ To find $A$ and $B$ , we multiply both sides by the denominator $(x-2)(x+5)$ to get: $\newline$ $7 = A(x+5) + B(x-2)$
Solve for Constants $A$ and $B$ : Solve for $A$ and $B$ by substituting suitable values for $x$ . Let $x = 2$ , then we get: $7 = A(2+5) + B(2-2)$ $7 = 7A$ $A = 1$ Let $x = -5$ , then we get: $B$ $0$ $B$ $1$ $B$ $2$
Rewrite Integral with Partial Fractions: Rewrite the integral with the found partial fractions. $\newline$ \int\frac{$7$}{(x$-2$)(x+$5$)}dx = \int\frac{$1$}{x$-2$}dx - \int\frac{$1$}{x+$5$}dx
Integrate Each Term Separately: Integrate each term separately. The integral of \(\frac{1}{(x-2)}dx is $\ln|x-2|$ , and the integral of $\frac{1}{(x+5)}$ dx is $\ln|x+5|$ . So, $\int\frac{1}{(x-2)}dx - \int\frac{1}{(x+5)}dx = \ln|x-2| - \ln|x+5| + C$ , where $C$ is the constant of integration.
Combine Results for Final Answer: Combine the results to write the final answer. $\newline$ The indefinite integral of $\frac{7}{(x-2)(x+5)}$ with respect to $x$ is $\ln|x-2| - \ln|x+5| + C$ .