Question

How do you integrate (3x^3 + 2x - 2)/(x^2 + 2) with respect to x? @Calculus1

Answer 1

long division

Answer 2

3x
----------------------
x^2+2 | 3x^3+2x-2
-(3x^3+6x)
---------------
-2x-2

Answer 3

\[\int\limits_{}^{}(3x+\frac{-2x-2}{x^2+2}) dx\]

Answer 4

\[3 \cdot \frac{x^{1+1}}{1+1} -\int\limits_{}^{}\frac{2x}{x^2+2} dx -\int\limits_{}^{}\frac{2}{x^2+2} dx\]

Answer 5

\[\text{ LET} u=x^2+2 => du=2x dx\]
\[3 \cdot \frac{x^2}{2}-\int\limits_{}^{}\frac{du}{u}-\int\limits_{}^{}\frac{2}{x^2+2} dx\]

\[\text{ Let } \tan(\theta)=\frac{x}{\sqrt{2}} => \sec^2(\theta) d \theta= \frac{1}{\sqrt{2}} dx\]
\[\frac{3}{2} x^2-\ln|u|-2 \cdot \int\limits_{}^{}\frac{1}{ (\sqrt{2} \tan(\theta))^2+2} \cdot \sqrt{2} \sec^2(\theta) d \theta\]
\[\frac{3}{2} x^2-\ln(x^2+2)-\frac{2}{2}\int\limits_{}^{}\frac{1}{\tan^2(\theta)+1} \cdot \sqrt{2} \sec^2(\theta) d \theta\]
\[\frac{3}{2}x^2-\ln(x^2+2)-\sqrt{2}\int\limits_{}^{} 1 d \theta\]
\[\frac{3}{2}x^2-\ln(x^2+2)-\sqrt{2} \theta+C\]
\[\frac{3}{2}x^2-\ln(x^2+2)-\sqrt{2}\tan^{-}(\frac{x}{\sqrt{2}})+C\]

Answer 6

Thank you