Question
$\text{Evaluate: } \int\frac{x^{2}}{1 + x^{3}} \text{dx}$
✓
Answer
$\text{Let} I \int\frac{x^{2}}{1 + x^{3}} \text{dx}$$\text{Putting 1} + x^{3} = t$
$\Rightarrow 3x^{2}\text{dx = dt}$
$\text{or x}^{2} dx = \frac{dt}{3} $
$\therefore I = \frac{1}{3} \int \frac{dt}{3} = \frac{1}{3}\log |t| + C $
$= \frac{1}{3} \log | 1 + x^{3}| + C$
$\Rightarrow 3x^{2}\text{dx = dt}$
$\text{or x}^{2} dx = \frac{dt}{3} $
$\therefore I = \frac{1}{3} \int \frac{dt}{3} = \frac{1}{3}\log |t| + C $
$= \frac{1}{3} \log | 1 + x^{3}| + C$
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