First slide

Methods of integration

Question

$\int \frac{x + x^{2 / 3} + x^{1 / 6}}{x (1 + x^{1 / 3})} d x =$

Difficult

A

$\frac{3}{2} x^{2 / 3} + 6 \tan^{- 1}$ $(x^{1 / 6}) + c$
B

$\frac{3}{2} x^{2 / 3} + \tan^{- 1}$ $(x^{1 / 6}) + c$
C

$\frac{3}{2} x^{2 / 3} + 6 \tan^{- 1}$ $x^{6} + c$
D

None of these

Solution

$I = \int \frac{x + x^{2 / 3} + x^{1 / 6}}{x (1 + x^{1 / 3})} d x$
L.C.M. of 3 and 6

$∴ x = t^{6}, d x = 6 t^{5} d t$

$∴ I = 6 \int \frac{(t^{6} + t^{4} + t) t^{5} d t}{t^{6} (1 + t^{2})} = 6 \int \frac{t^{5} + t^{3} + 1}{1 + t^{2}} d t$

$= 6 \int (t^{3} + \frac{1}{1 + t^{2}}) d t = 6 [\frac{t^{4}}{4} + \tan^{- 1} t] + c$

$=$ $\frac{3}{2} x^{2 / 3} + 6 \tan^{- 1}$ $(x^{1 / 6}) + c$ .

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