Question

How do you find the integral of `sqrt(x^2+9) dx`?

Answer 1

`-1/2*(t^2/2+18ln(t)-81/82t^3)+C` where `t=sqrt(x^2+9)-x`

Explanation:

Setting
`sqrt(x^2+9)=t+x`
then we get
`x=(9-t^2)/(2*t)`
and
`dx=-(t^2+9)/(2t^2)dt`
so we get`-1/2int (t^2+9)^2/t^3dt`
this is
`-1/2int (t+18/t+81/t^3)dt=`
`-1/2*(t^2/2+18ln(t)-81/(2t^2))+C`

Answer 2

We know that,

`intsqrt(x^2+a^2)dx=x/2sqrt(x^2+a^2)+a^2/2ln|x+sqrt(x^2+a^2)|+c`

Taking, `a=3` , we get

`intsqrt(x^2+9)dx=x/2sqrt(x^2+9)+9/2ln|x+sqrt(x^2+9)|+c`

Explanation:

`II^(nd) Method`

Here,

`I=sqrt(x^2+9) dx...to(1)`

`I=intsqrt(x^2+9)*1dx`

`"Using "color(blue)"Integration by Parts :"`

`color(blue)(intu*vdx=uintvdx-int(u'intvdx)dx`

Let `u=sqrt(x^2+9) and v=1`

`=>u'=(2x)/(2sqrt(x^2+9))=x/sqrt(x^2+9) and intvdx=x+c`

So,

`I=sqrt(x^2+9)*x-intx/sqrt(x^2+9)*xdx`

`I=xsqrt(x^2+9)-intx^2/sqrt(x^2+9)dx`

`I=xsqrt(x^2+9)-int((x^2+9)-9)/sqrt(x^2+9)dx`

`I=xsqrt(x^2+9)-int(x^2+9)/sqrt(x^2+9)dx+int9/sqrt(x^2+9)dx`

`I=xsqrt(x^2+9)-intsqrt(x^2+9)dx+9int1/sqrt(x^2+3^2)dx`

`I=xsqrt(x^2+9)-I+9ln|x+sqrt(x^2+3^2)|+Cto`from `(1)`

`I+I=xsqrt(x^2+9)+9ln|x+sqrt(x^2+9)|+C`

`2I=xsqrt(x^2+9)+9ln|x+sqrt(x^2+9)|+C`

`I=x/2sqrt(x^2+9)+9/2ln|x+sqrt(x^2+9)|+C`