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int_(0)^(1)(dx)/(4x^(2)+4x+5).

$9$ . $\int_{0}^{1} \frac{d x}{4 x^{2}+4 x+5}$ .

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Evaluate definite integrals using the chain rule

Full solution

Q.

9

.

\int_{0}^{1} \frac{d x}{4 x^{2}+4 x+5}

.

Rewrite integral with common factor: Rewrite the integral with a common factor in the denominator: $\newline$ $\int_{0}^{1} \frac{dx}{4x^2 + 4x + 5} = \int_{0}^{1} \frac{dx}{4(x^2 + x + \frac{5}{4})}$ .
Complete the square: Complete the square in the denominator: $\newline$ $x^2 + x + \frac{5}{4} = (x + \frac{1}{2})^2 + 1$ . $\newline$ So, $\int_{0}^{1} \frac{dx}{4(x^2 + x + \frac{5}{4})} = \int_{0}^{1} \frac{dx}{4((x + \frac{1}{2})^2 + 1)}$ .
Substitute $u$ and $du$ : Substitute $u = x + \frac{1}{2}$ , then $du = dx$ : $\newline$ When $x = 0$ , $u = \frac{1}{2}$ ; when $x = 1$ , $u = \frac{3}{2}$ . $\newline$ The integral becomes $\int_{\frac{1}{2}}^{\frac{3}{2}} \frac{du}{4(u^2 + 1)}$ .
Factor out constant: Factor out the constant from the integral: $\int_{1/2}^{3/2} \frac{du}{4(u^2 + 1)} = \frac{1}{4} \int_{1/2}^{3/2} \frac{du}{u^2 + 1}$ .
Recognize integral as arctan: Recognize the integral of $\frac{1}{u^2 + 1}$ as arctan( $u$ ): $\newline$ $\frac{1}{4} \int_{\frac{1}{2}}^{\frac{3}{2}} \frac{du}{u^2 + 1} = \frac{1}{4} [\text{arctan}(u)]$ from $\frac{1}{2}$ to $\frac{3}{2}$ .
Evaluate arctan at bounds: Evaluate the arctan function at the bounds: $\frac{1}{4} [\arctan(\frac{3}{2}) - \arctan(\frac{1}{2})]$ .
Calculate final result: Calculate the values of $\arctan(\frac{3}{2})$ and $\arctan(\frac{1}{2})$ : $\newline$ $\arctan(\frac{3}{2}) \approx 0.9828$ , $\arctan(\frac{1}{2}) \approx 0.4636$ . $\newline$ $\frac{1}{4} [0.9828 - 0.4636] = \frac{1}{4} \times 0.5192 = 0.1298$ .

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