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Integrate x3+xx2+2dx\int\frac{x^{3}+x}{x^{2}+2}\,dx.

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Q. Integrate x3+xx2+2dx\int\frac{x^{3}+x}{x^{2}+2}\,dx.
  1. Simplify Integrand: Simplify the integrand by dividing each term in the numerator by each term in the denominator.\newlineWe have the integral:\newlinex3+xx2+2dx\int\frac{x^3 + x}{x^2 + 2}\,dx\newlineWe can split this into two separate integrals:\newlinex3x2+2dx\int\frac{x^3}{x^2 + 2}\,dx + xx2+2dx\int\frac{x}{x^2 + 2}\,dx
  2. Split into Two Integrals: Simplify the first integral by recognizing that the numerator is the derivative of the denominator.\newlineFor the first integral, we have:\newlinex3x2+2dx\int\frac{x^3}{x^2 + 2}\,dx\newlineLet u=x2+2u = x^2 + 2, then du=2xdxdu = 2x\,dx, which means xdx=du2x\,dx = \frac{du}{2}.\newlineSubstitute uu and dudu into the integral:\newline12u2udu\frac{1}{2} \int\frac{u - 2}{u}\,du
  3. Substitute and Integrate: Simplify the second integral by recognizing it as a standard form.\newlineFor the second integral, we have:\newlinexx2+2dx\int\frac{x}{x^2 + 2}\,dx\newlineThis is a standard form that can be integrated directly.\newlineLet's integrate it:\newline12lnx2+2+C\frac{1}{2} \ln|x^2 + 2| + C
  4. Integrate First Integral: Integrate the first integral after substitution.\newlineWe have:\newline12(u2u)du\frac{1}{2} \int(\frac{u - 2}{u}) \, du\newlineSplit the integral:\newline12(1du21udu)\frac{1}{2} (\int 1\,du - 2 \int \frac{1}{u}\,du)\newlineIntegrate term by term:\newline12(u2lnu)+C\frac{1}{2} (u - 2\ln|u|) + C
  5. Substitute Back: Substitute back the original variable xx into the first integral.\newlineWe have:\newline12(u2lnu)+C\frac{1}{2} * (u - 2\ln|u|) + C\newlineSubstitute back u=x2+2u = x^2 + 2:\newline12((x2+2)2lnx2+2)+C\frac{1}{2} * ((x^2 + 2) - 2\ln|x^2 + 2|) + C
  6. Combine Final Answer: Combine the results from Step 33 and Step 55 to get the final answer.\newlineWe have:\newline12×((x2+2)2lnx2+2)+12×lnx2+2+C\frac{1}{2} \times ((x^2 + 2) - 2\ln|x^2 + 2|) + \frac{1}{2} \times \ln|x^2 + 2| + C\newlineSimplify the expression:\newline12×x2+1lnx2+2+C\frac{1}{2} \times x^2 + 1 - \ln|x^2 + 2| + C