Question

How do you find the integral of `sqrt(x²-49)dx`?

Answer 1

`I=sqrt(x^2-7^2)dx`
We know that,
`color(red)((1)intsqrt(x^2-a^2)dx=x/2sqrt(x^2-a^2)-a^2/2ln|x+sqrt(x^2-a^2)|+c`,
put `x=7`
`I=x/2sqrt(x^2-49)-49/2ln|x+sqrt(x^2-49)|+c`

Explanation:

We can solve above question `"without using"color(red)" Result(1)."`

Please see below.

`I=intsqrt(x^2-49)dx...to(A)`

`=int1*sqrt(x^2-49)dx`

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

`int(u*v)dx=uintvdx-int(u'intvdx)dx`

`u=sqrt(x^2-49) and v=1`

`=>u'=1/(2sqrt(x^2-49))2x=x/sqrt(x^2-49)andintvdx=x`

So,

`I=sqrt(x^2-49)*x-intx/sqrt(x^2-49)*xdx`

`=x*sqrt(x^2-49)-intx^2/sqrt(x^2-49)dx`

`=xsqrt(x^2-49)-int((x^2-49)+49)/sqrt(x^2-49)`

`I=xsqrt(x^2-49)-intsqrt(x^2-49)dx-int49/sqrt(x^2-49)dx`

`I=xsqrt(x^2-49)-I-49int1/sqrt(x^2-7^2)dx...toFrom(A)`

`I+I=xsqrt(x^2-49)-49ln|x+sqrt(x^2-7^2)|+c`

`2I=xsqrt(x^2-49)-49ln |x+sqrt(x^2-49)|+c`

`=>I=x/2sqrt(x^2-49)-49/2ln|x+sqrt(x^2-49)|+c`