Evaluate int(1)/(x^(2)+6x+13)dx (A) ln|(x+3)^(2)+4|+C (B) -(1)/(2)tan^(-1)((x+3)/(2))+C (C) -((1)/(x^(2)+6x+13))^(-2)+C (D) (1)/(2)tan^(-1)((x+3)/(2))+C

Simplify Denominator: Step 1: Simplify the denominator. We start by completing the square for the denominator x^2 + 6x + 13. (x + 3)^2 - 9 + 13 = (x + 3)^2 + 4.Rewrite Integral: Step 2: Rewrite the integral. Now, rewrite the integral using the completed square: int 1/((x + 3)^2 + 4) dx.Identify Integral Form: Step 3: Identify the integral form. This integral is of the form int 1/(u^2 + a^2) du, where u = x + 3 and a = 2.Use Standard Integral Formula: Step 4: Use the standard integral formula. The integral int 1/(u^2 + a^2) du = (1/a) * tan^(-1)(u/a) + C. Substitute u = x + 3 and a = 2: (1/2) * tan^(-1)((x + 3)/2) + C.

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Evaluate $\int\frac{1}{x^{2}+6x+13}\,dx$ $\newline$ (A) $\ln|\left(x+3\right)^{2}+4|+C$ $\newline$ (B) $-\frac{1}{2}\tan^{-1}\left(\frac{x+3}{2}\right)+C$ $\newline$ (C) $-\left(\frac{1}{x^{2}+6x+13}\right)^{-2}+C$ $\newline$ (D) $\frac{1}{2}\tan^{-1}\left(\frac{x+3}{2}\right)+C$

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Calculus

Evaluate definite integrals using the chain rule

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Q. Evaluate

\int\frac{1}{x^{2}+6x+13}\,dx

\newline

(A)

\ln|\left(x+3\right)^{2}+4|+C

\newline

(B)

-\frac{1}{2}\tan^{-1}\left(\frac{x+3}{2}\right)+C

\newline

(C)

-\left(\frac{1}{x^{2}+6x+13}\right)^{-2}+C

\newline

(D)

\frac{1}{2}\tan^{-1}\left(\frac{x+3}{2}\right)+C

Simplify Denominator: Step $1$ : Simplify the denominator. $\newline$ We start by completing the square for the denominator $x^2 + 6x + 13$ . $\newline$ $(x + 3)^2 - 9 + 13 = (x + 3)^2 + 4$ .
Rewrite Integral: Step $2$ : Rewrite the integral. $\newline$ Now, rewrite the integral using the completed square: $\newline$ $\int \frac{1}{(x + 3)^2 + 4} \, dx$ .
Identify Integral Form: Step $3$ : Identify the integral form. $\newline$ This integral is of the form $\int \frac{1}{u^2 + a^2} \, du$ , where $u = x + 3$ and $a = 2$ .
Use Standard Integral Formula: Step $4$ : Use the standard integral formula. $\newline$ The integral $\int \frac{1}{u^2 + a^2} \, du = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C$ . $\newline$ Substitute $u = x + 3$ and $a = 2$ : $\newline$ $\frac{1}{2} \tan^{-1}\left(\frac{x + 3}{2}\right) + C$ .