what is the integral of redical(x^2 +1)?

Question

please post the correct asnwer !! I will choose the best answer !!

it is the integration of redical of x squared plus 1

Answer 1

Let x = tanΘ.

Then x² = tan²Θ, and x² + 1 = tan²Θ + 1 = sec²Θ.
Also, √(x² + 1) = √(tan²Θ + 1) = √sec²Θ = secΘ .

So your integral is equivalent to ƒ secΘ dΘ =
ln |secΘ + tanΘ| +c, where c is a constant of integration.

Answer 2

[integral] sqrt(x^2 +1) dx =
[integral] (x^2 + 1 ) ^ 1/2 dx =
[ (x^2 + 1)^ 3/2] / 3/2 + C =
2/3 sqrt[ (x^2 +1) ^ 3 ] + C =
2/3 (x^2 +1) sqrt (x^2 +1) + C

Answer 3

integral(x^2+1)dx
= [ x^3/3 ] + x

formula is :
intergration(x^n)dx= x^ [(n+1)/(n+1)]
integration of dx = x

Answer 4

you need to do your own work. any calculus textbook should give this in a table. otherwise, your teacher should be able to help you. you will never learn if you have others do your work for you

Answer 5

substitution:
x=tanu
dx/du=sec^2u
dx=sec^2udu
S [√(tan^2u+1)]sec^2u du
=S (√sec^2u)sec^2u du
=S secusec^2u du
= secutanu - S (secutanu)tanu du
=secutanu-S secutan^2u du
I can't solve any further