Use integration by substitution to evaluate this definite integral.      ln 2 e^x√e^x + 4 dx

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To solve the integral \(\int_{\ln(2)}^{e^x} \sqrt{e^x + 4} \, dx\) using substitution, let's go through the steps.

### Step 1: Choose a substitution
We observe that the expression under the square root is \(e^x + 4\). This suggests a substitution that will simplify the integrand. Let:

\[
u = e^x + 4
\]

### Step 2: Differentiate \(u\)
Differentiate both sides with respect to \(x\):

\[
\frac{du}{dx} = e^x
\]

This implies:

\[
du = e^x \, dx
\]

### Step 3: Change the limits of integration
The original limits of integration are from \(x = \ln(2)\) to \(x = 0\). We need to convert these limits in terms of \(u\).

For \(x = \ln(2)\):

\[
u = e^{\ln(2)} + 4 = 2 + 4 = 6
\]

For \(x = 0\):

\[
u = e^0 + 4 = 1 + 4 = 5
\]

So the limits of integration in terms of \(u\) are from \(u = 6\) to \(u = 5\).

### Step 4: Rewrite the integral
The integral in terms of \(u\) becomes:

\[
\int_{6}^{5} \sqrt{u} \cdot \frac{du}{e^x}
\]

Since \(du = e^x \, dx\), the integral simplifies to:

\[
\int_{6}^{5} \sqrt{u} \, du
\]

### Step 5: Integrate with respect to \(u\)
Now, integrate \(\sqrt{u}\) with respect to \(u\):

\[
\int \sqrt{u} \, du = \int u^{1/2} \, du
\]

The antiderivative is:

\[
\frac{2}{3} u^{3/2} + C
\]

### Step 6: Evaluate the definite integral
Now, apply the limits of integration from 6 to 5:

\[
\left[ \frac{2}{3} u^{3/2} \right]_{6}^{5} = \frac{2}{3} \left( 5^{3/2} - 6^{3/2} \right)
\]

### Step 7: Simplify the expression
Now compute the values:

\[
5^{3/2} = \sqrt{5^3} = \sqrt{125} \approx 11.1803
\]
\[
6^{3/2} = \sqrt{6^3} = \sqrt{216} \approx 14.6969
\]

Thus, the integral is:

\[
\frac{2}{3} (11.1803 - 14.6969) = \frac{2}{3} \times -3.5166 \approx -2.3444
\]

So, the final result is approximately:

\[
\boxed{-2.3444}
\]

Would you like further details on any of the steps?

Learn how to evaluate the definite integral ln 2 e^x√e^x + 4 dx using integration by substitution. Follow a step-by-step solution with detailed explanations and calculations. Suitable for advanced high school and college students.

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