Please,someone help to solve the problem?

We know that `1+tan^2 u=sec^2u`

By the change proposed we have

`dx=sec^2u du`. Lets substitute in the integral

`intdx/(1+x^2)^(3/2)=intsec^2u/(1+tan^2u)^(3/2)du= intsec^2u/sec^3udu=int1/secudu=intcosudu=sinu+C`

Thus, undoing the change:

`u=arctanx` and finally we have

`sin u +C=sin(arctanx)+C`

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Let's try to use Trigonometric Substitution for solving this integral. To do so, we will construct a right angle triangle `Delta ABC` and label the sides in such a way that using Pythagoras' formula we can derive the expressions we are currently seeing in the argument of the integral as follows:

Angle `/_B=theta` has opposite side `x` and adjacent side `1`. Using Pythagoras' formula:

`(BC)^2=(AB)^2+(AC)^2` results in:

`(BC)^2=1^2+x^2=1+x^2`

`BC=sqrt(1+x^2` as shown.

Now, let's write the three most basic trigonometric functions for `theta`:

`sintheta=x/sqrt(1+x^2)`

`costheta=1/sqrt(1+x^2)`

`tantheta=x/1=x`

Now we need to use these equations to solve for various pieces of the integral argument in trigonometric terms. Let's use `tantheta`:

`tantheta=x`

Let's take derivatives of both sides:

`sec^2 theta d theta=dx`

From the `costheta` equation, we can solve for `sqrt(1+x^2)`:

`sqrt(1+x^2)=1/costheta=sectheta`

If we raise both sides of this equation to the power of `3` we get:

`sec^3theta=(sqrt(1+x^2))^3=((1+x^2)^(1/2))^3=(1+x^2)^(3/2)`

Now, we can substitute what we have calculated into the problem integral to turn it into a trigonometric integral:

`intdx/(1+x^2)^(3/2)=int(sec^2thetad theta)/sec^3theta=intsec^2theta/(secthetasec^2theta)d theta=intcancelcolor(red)(sec^2theta)/(secthetacancelcolor(red)(sec^2theta))d theta=int1/secthetad theta=int1/(1/costheta)d theta=intcosthetad theta=sintheta+C`

Now, we can substitute back for `sintheta` and turn our answer back into an algebraic expression in terms of `x`:

`color(blue)(intdx/(1+x^2)^(3/2)=x/sqrt(1+x^2)+C)`