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To solve the definite integral [tex]\(\int_1^2 \frac{\ln(x)}{x^2} \, dx\)[/tex], follow these steps:
1. Identify the integral and substitution: Notice that the integrand involves both [tex]\(\ln(x)\)[/tex] and a power of [tex]\(x\)[/tex]. A useful substitution would be [tex]\( u = \ln(x) \)[/tex]. Consequently, we have [tex]\( du = \frac{1}{x} \, dx \)[/tex]. Also, [tex]\(\frac{dx}{x} = du\)[/tex].
2. Change the limits of integration: With [tex]\(x = 1\)[/tex], we get [tex]\( u = \ln(1) = 0 \)[/tex]. With [tex]\(x = 2\)[/tex], we get [tex]\( u = \ln(2) \)[/tex]. So the integral limits change from [tex]\(x \in[1, 2]\)[/tex] to [tex]\(u \in[0, \ln(2)]\)[/tex].
3. Rewrite the integrand: The given integral becomes:
[tex]\[ \int_1^2 \frac{\ln(x)}{x^2} \, dx = \int_{\ln(1)}^{\ln(2)} \frac{u}{e^u} \, du \][/tex]
Here, we used the fact that [tex]\(x = e^u\)[/tex].
4. Simplify the integrand: Note that:
[tex]\[ \frac{u}{e^u} \text{ is a straightforward integration since } \int \frac{u}{e^u} \, du \][/tex]
We can use integration by parts where [tex]\( v = u \)[/tex] and [tex]\( dv = \frac{1}{e^u} \, du \)[/tex].
5. Apply integration by parts: Let [tex]\( u = u \)[/tex] and [tex]\( dv = \frac{1}{e^u} \, du \)[/tex]:
[tex]\[ \text{Choose } u = u \implies du = du \][/tex]
[tex]\[ \text{Choose } dv = \frac{1}{e^u} \implies v = \int \frac{1}{e^u} \, du = -\frac{1}{e^u} \][/tex]
So by the integration by parts formula, [tex]\(\int u \, dv = uv - \int v \, du\)[/tex]:
[tex]\[ \int \frac{u}{e^u} \, du = -\frac{u}{e^u} - \int -\frac{1}{e^u} \, du = -\frac{u}{e^u} + \int \frac{1}{e^u} \, du \][/tex]
Since [tex]\(\int \frac{1}{e^u} \, du = -\frac{1}{e^u}\)[/tex], we have:
[tex]\[ \int \frac{u}{e^u} \, du = -\frac{u}{e^u} - \frac{1}{e^u} \][/tex]
6. Evaluate the definite integral: Substitute back the limits [tex]\( u = 0 \)[/tex] to [tex]\( u = \ln(2) \)[/tex]:
[tex]\[ \left. -\frac{u}{e^u} - \frac{1}{e^u} \right|_{0}^{\ln(2)} \][/tex]
Calculate at the boundaries:
[tex]\[ \left( -\frac{\ln(2)}{e^{\ln(2)}} - \frac{1}{e^{\ln(2)}} \right) - \left( -\frac{0}{e^0} - \frac{1}{e^0} \right) \][/tex]
Simplify using [tex]\(e^{\ln(2)} = 2\)[/tex]:
[tex]\[ \left( -\frac{\ln(2)}{2} - \frac{1}{2} \right) - \left( 0 - 1 \right) \][/tex]
[tex]\[ -\frac{\ln(2)}{2} - \frac{1}{2} + 1 \][/tex]
Combine the terms:
[tex]\[ 1 - \left( \frac{1 + \ln(2)}{2} \right) \][/tex]
Which simplifies to:
[tex]\[ 1 - \frac{1}{2} - \frac{\ln(2)}{2} = \frac{1}{2} - \frac{\ln(2)}{2} \][/tex]
Therefore, the value of the definite integral [tex]\(\int_1^2 \frac{\ln(x)}{x^2} \, dx\)[/tex] is [tex]\(\frac{1}{2} - \frac{\ln(2)}{2}\)[/tex].
1. Identify the integral and substitution: Notice that the integrand involves both [tex]\(\ln(x)\)[/tex] and a power of [tex]\(x\)[/tex]. A useful substitution would be [tex]\( u = \ln(x) \)[/tex]. Consequently, we have [tex]\( du = \frac{1}{x} \, dx \)[/tex]. Also, [tex]\(\frac{dx}{x} = du\)[/tex].
2. Change the limits of integration: With [tex]\(x = 1\)[/tex], we get [tex]\( u = \ln(1) = 0 \)[/tex]. With [tex]\(x = 2\)[/tex], we get [tex]\( u = \ln(2) \)[/tex]. So the integral limits change from [tex]\(x \in[1, 2]\)[/tex] to [tex]\(u \in[0, \ln(2)]\)[/tex].
3. Rewrite the integrand: The given integral becomes:
[tex]\[ \int_1^2 \frac{\ln(x)}{x^2} \, dx = \int_{\ln(1)}^{\ln(2)} \frac{u}{e^u} \, du \][/tex]
Here, we used the fact that [tex]\(x = e^u\)[/tex].
4. Simplify the integrand: Note that:
[tex]\[ \frac{u}{e^u} \text{ is a straightforward integration since } \int \frac{u}{e^u} \, du \][/tex]
We can use integration by parts where [tex]\( v = u \)[/tex] and [tex]\( dv = \frac{1}{e^u} \, du \)[/tex].
5. Apply integration by parts: Let [tex]\( u = u \)[/tex] and [tex]\( dv = \frac{1}{e^u} \, du \)[/tex]:
[tex]\[ \text{Choose } u = u \implies du = du \][/tex]
[tex]\[ \text{Choose } dv = \frac{1}{e^u} \implies v = \int \frac{1}{e^u} \, du = -\frac{1}{e^u} \][/tex]
So by the integration by parts formula, [tex]\(\int u \, dv = uv - \int v \, du\)[/tex]:
[tex]\[ \int \frac{u}{e^u} \, du = -\frac{u}{e^u} - \int -\frac{1}{e^u} \, du = -\frac{u}{e^u} + \int \frac{1}{e^u} \, du \][/tex]
Since [tex]\(\int \frac{1}{e^u} \, du = -\frac{1}{e^u}\)[/tex], we have:
[tex]\[ \int \frac{u}{e^u} \, du = -\frac{u}{e^u} - \frac{1}{e^u} \][/tex]
6. Evaluate the definite integral: Substitute back the limits [tex]\( u = 0 \)[/tex] to [tex]\( u = \ln(2) \)[/tex]:
[tex]\[ \left. -\frac{u}{e^u} - \frac{1}{e^u} \right|_{0}^{\ln(2)} \][/tex]
Calculate at the boundaries:
[tex]\[ \left( -\frac{\ln(2)}{e^{\ln(2)}} - \frac{1}{e^{\ln(2)}} \right) - \left( -\frac{0}{e^0} - \frac{1}{e^0} \right) \][/tex]
Simplify using [tex]\(e^{\ln(2)} = 2\)[/tex]:
[tex]\[ \left( -\frac{\ln(2)}{2} - \frac{1}{2} \right) - \left( 0 - 1 \right) \][/tex]
[tex]\[ -\frac{\ln(2)}{2} - \frac{1}{2} + 1 \][/tex]
Combine the terms:
[tex]\[ 1 - \left( \frac{1 + \ln(2)}{2} \right) \][/tex]
Which simplifies to:
[tex]\[ 1 - \frac{1}{2} - \frac{\ln(2)}{2} = \frac{1}{2} - \frac{\ln(2)}{2} \][/tex]
Therefore, the value of the definite integral [tex]\(\int_1^2 \frac{\ln(x)}{x^2} \, dx\)[/tex] is [tex]\(\frac{1}{2} - \frac{\ln(2)}{2}\)[/tex].
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