Evaluate \( \int^{4} \frac{1}{x^{2}+2 x+3} d x \)

Evaluate the integral by following steps: - step0: Evaluate using substitution: \(\int_{0}^{4} \frac{1}{x^{2}+2x+3} dx\) - step1: Evaluate the integral: \(\int \frac{1}{x^{2}+2x+3} dx\) - step2: Rewrite the expression: \(\int \frac{1}{2+\left(x+1\right)^{2}} dx\) - step3: Use the substitution \(dx=1 dt\) to transform the integral\(:\) \(\int \frac{1}{2+\left(x+1\right)^{2}}\times 1 dt\) - step4: Simplify: \(\int \frac{1}{2+\left(x+1\right)^{2}} dt\) - step5: Use the substitution \(t=x+1\) to transform the integral\(:\) \(\int \frac{1}{2+t^{2}} dt\) - step6: Evaluate the integral: \(\frac{1}{\sqrt{2}}\times \arctan\left(\frac{t}{\sqrt{2}}\right)\) - step7: Substitute back: \(\frac{1}{\sqrt{2}}\times \arctan\left(\frac{x+1}{\sqrt{2}}\right)\) - step8: Simplify: \(\frac{\sqrt{2}}{2}\arctan\left(\frac{\sqrt{2}}{2}x+\frac{\sqrt{2}}{2}\right)\) - step9: Return the limits: \(\left(\frac{\sqrt{2}}{2}\arctan\left(\frac{\sqrt{2}}{2}x+\frac{\sqrt{2}}{2}\right)\right)\bigg |_{0}^{4}\) - step10: Calculate the value: \(\frac{\sqrt{2}\times \arctan\left(2\sqrt{2}+\frac{\sqrt{2}}{2}\right)-\sqrt{2}\times \arctan\left(\frac{\sqrt{2}}{2}\right)}{2}\) To evaluate the integral \[ \int^{4} \frac{1}{x^{2}+2 x+3} d x, \] we first need to simplify the integrand. The expression in the denominator can be rewritten by completing the square: \[ x^{2} + 2x + 3 = (x+1)^{2} + 2. \] Now, we can rewrite the integral as: \[ \int^{4} \frac{1}{(x+1)^{2} + 2} d x. \] Next, we can use the substitution \( u = x + 1 \), which gives \( du = dx \). The limits of integration change accordingly: when \( x = 0 \), \( u = 1 \) and when \( x = 4 \), \( u = 5 \). Thus, the integral becomes: \[ \int_{1}^{5} \frac{1}{u^{2} + 2} du. \] The integral \[ \int \frac{1}{u^{2} + a^{2}} du = \frac{1}{a} \arctan\left(\frac{u}{a}\right) + C, \] where \( a = \sqrt{2} \) in our case. Therefore, we have: \[ \int \frac{1}{u^{2} + 2} du = \frac{1}{\sqrt{2}} \arctan\left(\frac{u}{\sqrt{2}}\right) + C. \] Now we can evaluate the definite integral: \[ \left[ \frac{1}{\sqrt{2}} \arctan\left(\frac{u}{\sqrt{2}}\right) \right]_{1}^{5}. \] Calculating this gives: \[ \frac{1}{\sqrt{2}} \left( \arctan\left(\frac{5}{\sqrt{2}}\right) - \arctan\left(\frac{1}{\sqrt{2}}\right) \right). \] Thus, the final result for the integral is: \[ \frac{1}{\sqrt{2}} \left( \arctan\left(\frac{5}{\sqrt{2}}\right) - \arctan\left(\frac{1}{\sqrt{2}}\right) \right). \]