Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals2 Marks
Question
Find the integral: $\int \frac{x^{3}-x^{2}+x-1}{x-1} d x$
✓
Answer
$I = \int \frac{x^{3}-x^{2}+x-1}{x-1} d x$
Now the numerator can be factorized as,
$x^3 - x^2 + x - 1 = x^2(x - 1) + 1(x - 1)$
$x^3 - x^2 + x - 1 = (x^2 + 1)(x - 1)$
Now putting this in given integral we get,
$I = \frac{x^{3}-x^{2}+x-1}{x-1}=\frac{\left(x^{2}+1\right)(x-1)}{x-1}=x^{2}+1$
$= \int\left(x^{2}+1\right) d x$
$= \int x^{2} d x+\int 1 . d x$
$= \frac{x^{3}}{3}+x+C$
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