First slide

Methods of integration

Question

$If \int f (x) d x = F (x), then \int x^{2 n - 1} f (x^{n}) d x is equal to$

Moderate

A

$\frac{1}{n} [x^{n} F (x^{n})] - \int F (x^{n}) d (x^{n}) + C$
B

$\frac{1}{n} [x^{n} F (x^{n}) - \int F (x^{n}) d (x^{n})] + C$
C

$\frac{1}{n} [x^{n} F (x^{n})] - \int F (x^{n}) d (x) + C$
D

None of these

Solution

$I = \int x^{2 n - 1} f (x^{n}) d x = \frac{1}{n} \int x^{n} f (x^{n}) n x^{n - 1} d x$
$Putting y = x^{n}, we get$
$\begin{array}{l} \Rightarrow I = \frac{1}{n} \int yf (y) dy \\ = \frac{1}{n} [y \int f (y) dy - \int (1 \cdot \int f (y) dy) dy] \\ = \frac{1}{n} [yF (y) - \int F (y) dy] + C \\ = \frac{1}{n} [x^{n} F (x^{n}) - \int F (x^{n}) d (x^{n})] + C \end{array}$

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