00:01
All right, question 25.
00:08
We have indefinite integral of x plus 6 over the square root of x dx.
00:23
So what i did was i took x to the negative 1 half power times x plus 6.
00:36
And i further simplified that by distributing.
00:41
So x to the negative 1 half times x to the first, we add our exponents, and that becomes x to the positive 1 half, plus 6 times x to the negative 1 half, dx.
01:05
And then i integrated, so i add 1, so i have x to the 3 halves, multiplied by the reciprocal.
01:23
Plus 6x, add 1 to 1 half, multiplied by the reciprocal times 2, and add c...