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Answer
La integral es \( \sqrt{1+x^{2}}+3\ln{(x+\sqrt{1+x^{2}})} + C \).
Solution
Calculate the integral \( \int \frac{(x+3) dx}{\sqrt{1+x^{2}}} \).
Evaluate the integral by following steps:
- step0: Evaluate using formulas and rules:
\(\int \frac{\left(x+3\right)}{\sqrt{1+x^{2}}} dx\)
- step1: Remove the parentheses:
\(\int \frac{x+3}{\sqrt{1+x^{2}}} dx\)
- step2: Rewrite the fraction:
\(\int \left(\frac{x}{\sqrt{1+x^{2}}}+\frac{3}{\sqrt{1+x^{2}}}\right) dx\)
- step3: Use properties of integrals:
\(\int \frac{x}{\sqrt{1+x^{2}}} dx+\int \frac{3}{\sqrt{1+x^{2}}} dx\)
- step4: Evaluate the integral:
\(\left(1+x^{2}\right)^{\frac{1}{2}}+\int \frac{3}{\sqrt{1+x^{2}}} dx\)
- step5: Evaluate the integral:
\(\left(1+x^{2}\right)^{\frac{1}{2}}+3\ln{\left(x+\sqrt{1+x^{2}}\right)}\)
- step6: Simplify:
\(\sqrt{1+x^{2}}+3\ln{\left(x+\sqrt{1+x^{2}}\right)}\)
- step7: Add the constant of integral C:
\(\sqrt{1+x^{2}}+3\ln{\left(x+\sqrt{1+x^{2}}\right)} + C, C \in \mathbb{R}\)
La integral de \( \frac{(x+3) dx}{\sqrt{1+x^{2}}} \) es \( \sqrt{1+x^{2}}+3\ln{(x+\sqrt{1+x^{2}})} + C \), donde \( C \) es una constante real.
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Simplify this solution Extra Insights
To solve the integral \( \int \frac{(x+3) dx}{\sqrt{1+x^2}} \), we can split it into two simpler integrals: \[ \int \frac{x \, dx}{\sqrt{1+x^2}} + \int \frac{3 \, dx}{\sqrt{1+x^2}} \] The first integral, \( \int \frac{x \, dx}{\sqrt{1+x^2}} \), can be solved using the substitution \( u = 1 + x^2 \), which simplifies to \( \frac{1}{2} \sqrt{1+x^2} + C \). For the second integral, \( \int \frac{3 \, dx}{\sqrt{1+x^2}} \), this results in \( 3 \ln |x + \sqrt{1+x^2}| + C \). Combining both parts, we have: \[ \int \frac{(x+3) dx}{\sqrt{1+x^2}} = \sqrt{1+x^2} + 3 \ln |x + \sqrt{1+x^2}| + C \]
