What is the antiderivative of F(x) = x/ sqrt (x^2 + 1)F(x)=x√x2+1?

2 Answers
Sonnhard
May 31, 2018

sqrt(x^2+1)+C√x2+1+C

Explanation:

Writing your integral in the form

1/2*int2x/sqrt(x^2+1)dx12⋅∫2x√x2+1dx

Now we Substitute

x^2+1=tx2+1=t
then

2xdx=dt2xdx=dt

and our integral is
1/2int dt/sqrt(t)12∫dt√t
1/2*int t^{-1/2}dt=t^(1/2)+C12⋅∫t−12dt=t12+C
So we obtain

sqrt(x^2+1)+C√x2+1+C

maganbhai P.
May 31, 2018

Note: color(blue)(int(f'(x))/sqrt(f(x))dx=2sqrt(f(x))+c

I=intx/sqrt(x^2+1)dx=1/2int(2x)/sqrt(x^2+1)dx=1/2xx2sqrt(x^2+1)+c

:.I=sqrt(x^2+1)+c

Explanation:

II^(nd) Method :

I=intx/sqrt(x^2+1)dx

Let,

sqrt(x^2+1)=u=>x^2+1=u^2=>2xdx=2udu=>xdx=udu

So,

I=intu/sqrt(u^2)du=intu/udu=int1du=u+c

Subst. u=sqrt(x^2+1)

=>I=sqrt(x^2+1)+c

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