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

1 Answer
maganbhai P.
Apr 22, 2018

#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#

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