Evaluate the integral and express your answer in simplest form. int(-10)/(16+x^(2))dx Answer:

Recognize Standard Form: We are given the integral: int(-10)/(16+x^(2))dx To solve this integral, we can recognize that it is a standard form of the arctangent function, where the integral of 1/(a^2 + x^2)dx is (1/a)arctan(x/a) + C, where C is the constant of integration. Let's identify a^2 = 16, which gives us a = 4.Identify Constant: Now we can rewrite the integral by factoring out the constant -10 and substituting a = 4: int(-10)/(16+x^(2))dx = -10 * int(1)/(16+x^(2))dx = -10 * int(1)/(4^2+x^(2))dxRewrite Integral: Next, we apply the arctangent formula: -10 * int(1)/(4^2+x^(2))dx = -10 * (1/4)arctan(x/4) + C = (-10/4) * arctan(x/4) + C = -5/2 * arctan(x/4) + CApply Arctangent Formula: We have found the indefinite integral in its simplest form: -5/2 * arctan(x/4) + C

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Evaluate the integral and express your answer in simplest form.

int(-10)/(16+x^(2))dx
Answer:

Evaluate the integral and express your answer in simplest form. $\newline$ $\int \frac{-10}{16+x^{2}} d x$ $\newline$ Answer:

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Calculus

Find indefinite integrals using the substitution

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Q. Evaluate the integral and express your answer in simplest form.

\newline

\int \frac{-10}{16+x^{2}} d x

\newline

Answer:

Recognize Standard Form: We are given the integral: $\newline$ $\int \frac{-10}{16+x^{2}}\,dx$ $\newline$ To solve this integral, we can recognize that it is a standard form of the arctangent function, where the integral of $\frac{1}{a^{2} + x^{2}}\,dx$ is $\frac{1}{a}$ arctan $\frac{x}{a}$ + C, where C is the constant of integration. $\newline$ Let's identify $a^{2} = 16$ , which gives us $a = 4$ .
Identify Constant: Now we can rewrite the integral by factoring out the constant $-10$ and substituting $a = 4$ : $\newline$ $\int\frac{-10}{16+x^{2}}dx = -10 \times \int\frac{1}{16+x^{2}}dx$ $\newline$ $= -10 \times \int\frac{1}{4^2+x^{2}}dx$
Rewrite Integral: Next, we apply the arctangent formula: $\newline$ \(-10 \times \int \frac{ $1$ }{ $4$ ^ $2$ +x^{ $2$ }}dx = $-10$ \times \left(\frac{ $1$ }{ $4$ }\right)\arctan\left(\frac{x}{ $4$ }\right) + C $\newline$ = \left(-\frac{ $10$ }{ $4$ }\right) \times \arctan\left(\frac{x}{ $4$ }\right) + C $\newline$ = -\frac{ $5$ }{ $2$ } \times \arctan\left(\frac{x}{ $4$ }\right) + C
Apply Arctangent Formula: We have found the indefinite integral in its simplest form: $-\frac{5}{2} \cdot \arctan\left(\frac{x}{4}\right) + C$