How do you integrate #int (x^2-1)/sqrt(x^2+9)dx# using trigonometric substitution?

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Answer 1 · 2018-08-05T13:53:26

#int(x^2-1)/sqrt(x^2+9)dx=x/2sqrt(x^2+9)-11/2ln|x+sqrt(x^2+9)|+C#

Here ,

#int(x^2-1)/sqrt(x^2+9)dx=int(x^2+9-10)/sqrt(x^2+9)dx#

#int(x^2-1)/sqrt(x^2+9)dx=intsqrt(x^2+9)dx-10int1/sqrt(x^2+9)dx#

#int(x^2-1)/sqrt(x^2+9)dx=I-10ln|x+sqrt(x^2+9)|color(red)(...to(A)#

Now, #I=intsqrt(x^2+9)dx#

#:.I=intsqrt(x^2+9)*1dx#

#I=sqrt(x^2+9)int1dx-int(1/(2sqrt(x^2+9))(2x)int1dx)dx#

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

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

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

#I=xsqrt(x^2+9)-intsqrt(x^2+9)dx+int9/sqrt(x^2+9)dx#

#I=xsqrt(x^2+9)-I+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/2#

From eqn #color(red)((A)# we have

#int(x^2-1)/sqrt(x^2+9)dx=x/2sqrt(x^2+9)+9/2ln|x+sqrt(x^2+9)|#

#color(white)(............................................)-10ln|x+sqrt(x^2+9)|+c/2#

#int(x^2-1)/sqrt(x^2+9)dx=x/2sqrt(x^2+9)-11/2ln|x+sqrt(x^2+9)|+C#