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How do you evaluate: indefinite integral #(1+x)/(1+x^2) dx#?

Answer 1

Christopher Agresta

The answer is #=arctanx+1/2ln(1+x^2)+C#

The integral is

#I=int((1+x)dx)/(1+x^2)=int(dx)/(1+x^2)+int(xdx)/(1+x^2)#

The first integral

#int(dx)/(1+x^2)=arctanx#

And the second integral is

#int(xdx)/(1+x^2)=1/2int(2xdx)/(1+x^2)#

#=1/2ln(1+x^2)#

And finally

#I=arctanx+1/2ln(1+x^2)+C#

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Answer 2

Paisley Johnson

To evaluate the indefinite integral of (1+x)/(1+x^2) dx, you can use partial fraction decomposition followed by integration. After decomposing (1+x)/(1+x^2) into partial fractions, you integrate each term separately.

First, decompose (1+x)/(1+x^2) into partial fractions:

(1+x)/(1+x^2) = A/(1+x) + Bx/(1+x^2)

Solve for A and B by finding a common denominator:

1 + x = A(1+x^2) + Bx

Expand and equate coefficients:

1 + x = A + Ax^2 + Bx

Solve for A and B:

A = 1 B = 1

Now integrate each term separately:

∫(1+x)/(1+x^2) dx = ∫(1/(1+x) + x/(1+x^2)) dx

= ∫(1/(1+x)) dx + ∫(x/(1+x^2)) dx

= ln|1+x| + (-1/2)ln(1+x^2) + C

Where C is the constant of integration.

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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.