If n is an odd positive integer, then `int|x^(n)|dx` is equal to

Options

A. `|(x^(n+1))/(n+1)|+C`
B. `(x^(n+1))/(n+1)+C`
C. `(|x^(n)|)/(n+1)+C`
D. none of these

Answer

c

Explanation

We have the following cases :
CASE I When `xge0`
In this case , we have , `int|x^(n)|dx=int|x|^(n)dx`
`rArrint|x^(n)|dx=intx^(n)dx " " [because|x|=x]`
`rArrint|x^(n)|dx=(x^(n+1))/(n+1)+C`
`rArrint|x^(n)|dx=(|x|^(n)x)/(n+1)+C " " [becausexge0therefore|x|=x]`
CASE II When ` xle0`
In this case , we have
`|x| =-x`
`thereforeint|x^(n)|dx= int|x|^(n)dx=int (-x)^(n)dx`
`rArrint|x^(n)|dx=-intx^(n)dx " " [because n is odd"]`
`rArrint|x^(n)|dx=-(x^(n)+1)/(n+1)+C`
`rArr int|x^(n)|dx=((-x)^(n)x)/(n+1)+C " " [{:(because " n is odd",),(therefore(-x)^(n)=-x^(n),):}]`
`rArrint|x^(n)|dx=(|x|^(n)x)/(n+1)+C`
Hence , `int|x^(n)|dx=(|x|^(n)x)/(n+1)+C`