00:01
Hi, now we are going to integrate the following.
00:04
The first question is integral over 3x divided by square root of 1 plus 3x squared dx.
00:15
Now we take u is equal to 3x squared plus 1.
00:20
Then d u is equal to 6x into dx.
00:25
Then the given integral is equal to 1 by 2 integral over 3x 1 by root u into d u and it can be written as 1 by 2 into 2 root u plus c and it will be equal to root u plus c then it can be written as root of 3 x squared plus 1 plus c therefore the answer for the first question is root of 3 x squared plus 1 plus c.
01:11
Now the second question is integral over 2x minus 1 the whole squared into dx.
01:20
Now let us take u equal to 4x, then du is equal to 4dx, then the given integral can be written as integral over 2cun u minus 1 the whole square du.
01:38
It will be equal to tan u minus 2 ln of tan u plus second u plus u plus u plus c integral over second 4x minus 1 the whole squared inter d x is equal to tan 4x minus 2 ln of tan 4x tan 4x minus 2 ln of tan 4 4 4x plus 2x 2x plus 4x plus 4x plus c and the third question is integral over d x divide by 1 plus 5x plus 1 the whole square in let us take u equal to 5x plus 1 then d u is equal to 5 d x x then the given integral can be written as 1 1 by 5 integral over 1 by u squared plus 1 into du after integration it will be equal to 1 by 5 10 inverse of u plus c therefore integral over d x divide by 1 plus 5 x divide by 1 plus 5 x plus 1 the whole square is equal to 1 by 5 5 tan inverse of 5x plus 1 plus c.
03:22
And next, the fourth question is integral over dx divide by x squared into root of x squared minus 1.
03:32
Now we take x equal to second u...