Question

Evaluate the Integral.

Answer 1

\[\int\limits_{}^{}\sqrt{x^2+2x}\]

Answer 2

dont forget the dx :)

Answer 3

okay do u know ho to deal with sqrt(x^2+1)

Answer 4

What do you mean?

Answer 5

if it was another intgegral sqrt(X^2 + 1) dx

Answer 6

you can do tan^2(u) sub

Answer 7

Oh yeah and then change it to sec right?

Answer 8

yes

Answer 9

so first u need to put it in a form similar to sqrt(x^2+1)

Answer 10

complete the square

Answer 11

use this
tan^2(u)+1=sec^2(u)

x^2+2x = (x+1)^2-2
we can use sec^2(u) -1 = tan^2(u)

Answer 12

So would i set x= tan(u) or to tan^2(u)?

Answer 13

sec u since its its in the form sqrt(x^2-1)

Answer 14

Why is it x^2 - 1 and not x^2 +1?

Answer 15

beause you rewrote
x^2+2x = (x+1)^2-1

Answer 16

y= x+1
sqrt(y^2 -1) dy
u=sec y
du=tan u sec u dy
integral sqrt(sec^2(u) -1) du
integral sqrt(tan^2(u))*tan u sec u du
integral tan^2(u)*sec(u) du

Answer 17

So we would get \[\int\limits_{}^{}\sqrt{(\sec(u)+1)^2-1} \sec(u)\tan(u)\]

Answer 18

the firs integraal line should read
integral sqrt(sec^2(u) -1) *tan u sec u du

Answer 19

no we completed the squre so we can sub everything in the bracket away

Answer 20

u need it in form sqrt(sec^2(u) - 1 )

Answer 21

How does it become just sec^2 (u) -1?

Answer 22

Oh so you set y=x+1 got it.

Answer 23

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Answer 24

ya an since dy = dx still in this case no need to change anything with the dx

Answer 25

So then you would set w=sec(u) and dw= tan^2 (u) and then solve the integral to get,
w^2/2
sec^2(u)/2
sec^2(sec^-1(y))
sec^2(sec^-1(x+1))