How do you evaluate the integral of #int x/[(x^2)+1]dx#?

Answer 1

Note that the derivative of the denominator is almost the numerator. Use substitution.

#int x/[(x^2)+1]dx#
Let #u=x^2+1#, so that #du = 2xdx#.

The integral is

#int x/[(x^2)+1]dx = int 1/(x^2+1) xdx#
# = 1/2 int underbrace(1/(x^2+1))_(1/u) underbrace(2xdx)_(du)#
# = 1/2 int 1/u du#
# = 1/2 ln abs u +C#
# = 1/2 ln abs(x^2+1)+C#
# = 1/2 ln(x^2+1)+C#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To evaluate the integral ∫(x/(x^2 + 1)) dx, you can use a trigonometric substitution. Let ( u = x^2 + 1 ), so ( du = 2x dx ). This substitution allows us to express the integral in terms of ( u ), which simplifies the integration.

Substitute ( u = x^2 + 1 ), which implies ( du = 2x dx ). Thus, ( dx = \frac{du}{2x} ).

Now, rewrite the integral in terms of ( u ):

[ \int \frac{x}{x^2 + 1} dx = \frac{1}{2} \int \frac{1}{u} du ]

Integrating ( \frac{1}{u} ) with respect to ( u ), we get:

[ \frac{1}{2} \ln|u| + C ]

Substitute back for ( u ):

[ \frac{1}{2} \ln|x^2 + 1| + C ]

So, the integral of ( \frac{x}{x^2 + 1} dx ) is ( \frac{1}{2} \ln|x^2 + 1| + C ), where ( C ) is the constant of integration.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7