Question
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b) \( \int \frac{\sqrt{x}-1}{1+\sqrt{x}} \quad, t=\sqrt{x} \)

Ask by Mcguire Lindsey. in Spain
Dec 08,2024

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Answer

Para resolver la integral \( \int \frac{\sqrt{x}-1}{1+\sqrt{x}} \, dx \) con la sustitución \( t = \sqrt{x} \), se obtiene \( x - 2 - 2\sqrt{x} + 2\ln|1+\sqrt{x}| + C \).

Solution

Para resolver la integral \( \int \frac{\sqrt{x}-1}{1+\sqrt{x}} \, dx \) usando la sustitución \( t = \sqrt{x} \), primero debemos expresar \( dx \) en términos de \( dt \). Dado que \( t = \sqrt{x} \), entonces \( x = t^2 \) y derivando ambos lados con respecto a \( t \) obtenemos: \[ dx = 2t \, dt \] Ahora sustituimos \( \sqrt{x} \) y \( dx \) en la integral: \[ \int \frac{\sqrt{x}-1}{1+\sqrt{x}} \, dx = \int \frac{t-1}{1+t} \cdot 2t \, dt \] Simplificamos la integral: \[ = 2 \int \frac{t(t-1)}{1+t} \, dt \] Descomponemos la fracción: \[ = 2 \int \left( t - \frac{t}{1+t} \right) dt \] Ahora, resolvemos cada parte de la integral: 1. La primera parte: \[ 2 \int t \, dt = 2 \cdot \frac{t^2}{2} = t^2 \] 2. La segunda parte: \[ -2 \int \frac{t}{1+t} \, dt \] Para resolver \( \int \frac{t}{1+t} \, dt \), podemos usar la sustitución \( u = 1 + t \), lo que implica que \( du = dt \) y \( t = u - 1 \): \[ \int \frac{t}{1+t} \, dt = \int \frac{u-1}{u} \, du = \int \left( 1 - \frac{1}{u} \right) du = u - \ln|u| + C = (1+t) - \ln|1+t| + C \] Por lo tanto, tenemos: \[ -2 \int \frac{t}{1+t} \, dt = -2 \left( (1+t) - \ln|1+t| \right) = -2(1+t) + 2\ln|1+t| \] Ahora, combinamos ambas partes: \[ \int \frac{\sqrt{x}-1}{1+\sqrt{x}} \, dx = t^2 - 2(1+t) + 2\ln|1+t| + C \] Finalmente, sustituimos \( t = \sqrt{x} \): \[ = (\sqrt{x})^2 - 2(1+\sqrt{x}) + 2\ln|1+\sqrt{x}| + C \] Simplificando: \[ = x - 2 - 2\sqrt{x} + 2\ln|1+\sqrt{x}| + C \] Por lo tanto, la solución de la integral es: \[ \int \frac{\sqrt{x}-1}{1+\sqrt{x}} \, dx = x - 2 - 2\sqrt{x} + 2\ln|1+\sqrt{x}| + C \]

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The Deep Dive

To solve the integral \( \int \frac{\sqrt{x}-1}{1+\sqrt{x}} \, dx \) using the substitution \( t = \sqrt{x} \), first, we have \( x = t^2 \) and \( dx = 2t \, dt \). Substitute these into the integral: \[ \int \frac{t-1}{1+t} \cdot 2t \, dt = 2 \int \frac{t(t-1)}{1+t} \, dt \] Now simplify the integrand: \[ \frac{t(t-1)}{1+t} = \frac{t^2 - t}{1+t} \] This can be split into two separate integrals: \[ 2 \left( \int t \, dt - \int \frac{t}{1+t} \, dt \right) \] Solving the first integral simply yields: \[ 2 \cdot \frac{t^2}{2} = t^2 \] For the second integral, use the substitution \( u = 1 + t \), which leads to \( dt = du \): \[ \int \frac{t}{1+t} \, dt = \int \frac{u-1}{u} \, du = \int (1 - \frac{1}{u}) \, du = u - \ln |u| + C \] Substituting back \( t \) in place of \( u \): \[ = (1+t) - \ln |1+t| \] Thus, combining everything, we have: \[ 2 \left( t^2 - \left( (1+t) - \ln |1+t| \right) \right) + C \] Now substituting back \( t = \sqrt{x} \) gives the final answer in \( x \): \[ 2 \left( x - (1 + \sqrt{x}) + \ln |1 + \sqrt{x}| \right) + C \] So the integral evaluates to: \[ 2x - 2 - 2\sqrt{x} + 2\ln |1+\sqrt{x}| + C \]

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