Answer
Aquí están las soluciones a las integrales:
1. \( \int x \ln x \, dx = \frac{1}{2}x^{2}\ln{(x)} - \frac{x^{2}}{4} + C \)
2. \( \int x e^{3 x} \, dx = \frac{e^{3x}x}{3} - \frac{e^{3x}}{9} + C \)
3. \( \int x^{3} e^{x^{2}} \, dx = \frac{e^{x^{2}}x^{2}}{2} - \frac{e^{x^{2}}}{2} + C \)
4. \( \int \cos \sqrt{x} \, dx \) no se puede calcular con el método de sustitución.
5. \( \int \frac{x e^{x}}{(x+1)^{2}} \, dx = -\frac{xe^{x}}{x+1} + e^{x} + C \)
Si necesitas ayuda con la integral que no se pudo calcular o cualquier otra pregunta, avísame.
Solution
Evaluate the integral by following steps:
- step0: Evaluate using partial integration formula:
\(\int \frac{xe^{x}}{\left(x+1\right)^{2}} dx\)
- step1: Evaluate the integral:
\(\int \left(x+1\right)^{-2}xe^{x} dx\)
- step2: Prepare for integration by parts:
\(\begin{align}&u=xe^{x}\\&dv=\left(x+1\right)^{-2}dx\end{align}\)
- step3: Calculate the derivative:
\(\begin{align}&du=\left(e^{x}+xe^{x}\right)dx\\&dv=\left(x+1\right)^{-2}dx\end{align}\)
- step4: Evaluate the integral:
\(\begin{align}&du=\left(e^{x}+xe^{x}\right)dx\\&v=\frac{\left(x+1\right)^{-1}}{-1}\end{align}\)
- step5: Substitute the values into formula:
\(xe^{x}\times \frac{\left(x+1\right)^{-1}}{-1}-\int \left(e^{x}+xe^{x}\right)\times \frac{\left(x+1\right)^{-1}}{-1} dx\)
- step6: Calculate:
\(-xe^{x}\left(x+1\right)^{-1}-\int -e^{x} dx\)
- step7: Evaluate the integral:
\(-xe^{x}\left(x+1\right)^{-1}+e^{x}\)
- step8: Multiply the terms:
\(-\frac{xe^{x}}{x+1}+e^{x}\)
- step9: Add the constant of integral C:
\(-\frac{xe^{x}}{x+1}+e^{x} + C, C \in \mathbb{R}\)
Calculate the integral \( \int x \ln x \, dx \).
Evaluate the integral by following steps:
- step0: Evaluate using partial integration formula:
\(\int x\ln{\left(x\right)} dx\)
- step1: Prepare for integration by parts:
\(\begin{align}&u=\ln{\left(x\right)}\\&dv=xdx\end{align}\)
- step2: Calculate the derivative:
\(\begin{align}&du=\frac{1}{x}dx\\&dv=xdx\end{align}\)
- step3: Evaluate the integral:
\(\begin{align}&du=\frac{1}{x}dx\\&v=\frac{x^{2}}{2}\end{align}\)
- step4: Substitute the values into formula:
\(\ln{\left(x\right)}\times \frac{x^{2}}{2}-\int \frac{1}{x}\times \frac{x^{2}}{2} dx\)
- step5: Calculate:
\(\frac{x^{2}\ln{\left(x\right)}}{2}-\int \frac{x}{2} dx\)
- step6: Evaluate the integral:
\(\frac{x^{2}\ln{\left(x\right)}}{2}-\frac{x^{2}}{4}\)
- step7: Simplify the expression:
\(\frac{1}{2}x^{2}\ln{\left(x\right)}-\frac{x^{2}}{4}\)
- step8: Add the constant of integral C:
\(\frac{1}{2}x^{2}\ln{\left(x\right)}-\frac{x^{2}}{4} + C, C \in \mathbb{R}\)
Calculate the integral \( \int x e^{3 x} \, dx \).
Evaluate the integral by following steps:
- step0: Evaluate using partial integration formula:
\(\int xe^{3x} dx\)
- step1: Prepare for integration by parts:
\(\begin{align}&u=x\\&dv=e^{3x}dx\end{align}\)
- step2: Calculate the derivative:
\(\begin{align}&du=dx\\&dv=e^{3x}dx\end{align}\)
- step3: Evaluate the integral:
\(\begin{align}&du=dx\\&v=\frac{e^{3x}}{3}\end{align}\)
- step4: Substitute the values into formula:
\(x\times \frac{e^{3x}}{3}-\int 1\times \frac{e^{3x}}{3} dx\)
- step5: Calculate:
\(\frac{xe^{3x}}{3}-\int \frac{e^{3x}}{3} dx\)
- step6: Evaluate the integral:
\(\frac{xe^{3x}}{3}-\frac{e^{3x}}{9}\)
- step7: Simplify the expression:
\(\frac{e^{3x}x}{3}-\frac{e^{3x}}{9}\)
- step8: Add the constant of integral C:
\(\frac{e^{3x}x}{3}-\frac{e^{3x}}{9} + C, C \in \mathbb{R}\)
Calculate the integral \( \int x^{3} e^{x^{2}} \, dx \).
Evaluate the integral by following steps:
- step0: Evaluate using substitution:
\(\int x^{3}e^{\left(x^{2}\right)} dx\)
- step1: Use the substitution \(dx=\frac{1}{2x} dt\) to transform the integral\(:\)
\(\int x^{3}e^{\left(x^{2}\right)}\times \frac{1}{2x} dt\)
- step2: Simplify:
\(\int \frac{e^{\left(x^{2}\right)}x^{2}}{2} dt\)
- step3: Use the substitution \(t=x^{2}\) to transform the integral\(:\)
\(\int \frac{e^{t}t}{2} dt\)
- step4: Rewrite the expression:
\(\int \frac{1}{2}e^{t}t dt\)
- step5: Use properties of integrals:
\(\frac{1}{2}\times \int e^{t}t dt\)
- step6: Prepare for integration by parts:
\(\begin{align}&u=t\\&dv=e^{t}dt\end{align}\)
- step7: Calculate the derivative:
\(\begin{align}&du=dt\\&dv=e^{t}dt\end{align}\)
- step8: Evaluate the integral:
\(\begin{align}&du=dt\\&v=e^{t}\end{align}\)
- step9: Substitute the values into formula:
\(\frac{1}{2}\left(te^{t}-\int 1\times e^{t} dt\right)\)
- step10: Calculate:
\(\frac{1}{2}\left(te^{t}-\int e^{t} dt\right)\)
- step11: Calculate:
\(\frac{1}{2}te^{t}-\frac{1}{2}\times \int e^{t} dt\)
- step12: Evaluate the integral:
\(\frac{1}{2}te^{t}-\frac{e^{t}}{2}\)
- step13: Substitute back:
\(\frac{1}{2}x^{2}e^{\left(x^{2}\right)}-\frac{e^{\left(x^{2}\right)}}{2}\)
- step14: Rewrite the expression:
\(\frac{e^{\left(x^{2}\right)}x^{2}}{2}-\frac{e^{\left(x^{2}\right)}}{2}\)
- step15: Add the constant of integral C:
\(\frac{e^{\left(x^{2}\right)}x^{2}}{2}-\frac{e^{\left(x^{2}\right)}}{2} + C, C \in \mathbb{R}\)
Aquí están las soluciones a las integrales solicitadas:
1. Para \( \int x \ln x \, dx \):
\[
\frac{1}{2}x^{2}\ln{(x)} - \frac{x^{2}}{4} + C, \quad C \in \mathbb{R}
\]
2. Para \( \int x e^{3 x} \, dx \):
\[
\frac{e^{3x}x}{3} - \frac{e^{3x}}{9} + C, \quad C \in \mathbb{R}
\]
3. Para \( \int x^{3} e^{x^{2}} \, dx \):
\[
\frac{e^{(x^{2})}x^{2}}{2} - \frac{e^{(x^{2})}}{2} + C, \quad C \in \mathbb{R}
\]
4. Para \( \int \cos \sqrt{x} \, dx \), no se pudo calcular debido a que el método de sustitución no es compatible con esta integral.
5. Para \( \int \frac{x e^{x}}{(x+1)^{2}} \, dx \):
\[
-\frac{xe^{x}}{x+1} + e^{x} + C, \quad C \in \mathbb{R}
\]
Si necesitas más ayuda con la integral que no se pudo calcular o cualquier otra consulta, házmelo saber.
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Explain
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Simplify this solution