How do you integrate int (x^2-1)/sqrt(x^2+9)dx∫x2−1√x2+9dx using trigonometric substitution?

1 Answer
maganbhai P.
Aug 5, 2018

int(x^2-1)/sqrt(x^2+9)dx=x/2sqrt(x^2+9)-11/2ln|x+sqrt(x^2+9)|+C∫x2−1√x2+9dx=x2√x2+9−112ln∣∣x+√x2+9∣∣+C

Explanation:

Here ,

int(x^2-1)/sqrt(x^2+9)dx=int(x^2+9-10)/sqrt(x^2+9)dx∫x2−1√x2+9dx=∫x2+9−10√x2+9dx

int(x^2-1)/sqrt(x^2+9)dx=intsqrt(x^2+9)dx-10int1/sqrt(x^2+9)dx∫x2−1√x2+9dx=∫√x2+9dx−10∫1√x2+9dx

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

Using Integration by parts:

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

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