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Evaluate the indefinite integral:
[tex]\mathsf{\displaystyle\int\! \frac{x^2}{x^3+1}\,dx}\\\\\\
\mathsf{=\displaystyle\int\! \frac{1}{3}\cdot 3\cdot \frac{x^2}{x^3+1}\,dx}\\\\\\
\mathsf{=\displaystyle \frac{1}{3}\int\! \frac{1}{x^3+1}\cdot 3x^2\,dx\qquad\quad(i)}[/tex]
Make a substitution:
[tex]\mathsf{x^3+1=u\quad \Rightarrow\quad 3x^2\,dx=du}[/tex]
and the integral (i) becomes
[tex]\mathsf{=\displaystyle \frac{1}{3}\int\! \frac{1}{u}\,du}\\\\\\
\mathsf{=\displaystyle \frac{1}{3}\cdot \ell n\!\left|u\right|+C}[/tex]
Substitute back for u = x³ + 1, and you get the result:
[tex]\mathsf{=\dfrac{1}{3}\,\ell n\!\left|x^3+1\right|+C}[/tex]
[tex]\therefore~~\mathsf{\displaystyle\int\!\frac{x^2}{x^3+1}\,dx=\frac{1}{3}\,\ell n\!\left|x^3+1\right|+C}\qquad\quad\checkmark[/tex]
I hope this helps. =)
Tags: indefinite integral rational function substitution natural logarithm log ln differential integral calculus
