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)#