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How do you find the antiderivative of #(5x^2)/(x^2 + 1)#?

Answer 1

Emily Ammerman

I would start by factorizing out #x^2# from the denominator:

#int(5x^2)/(x^2+1)dx=int(5x^2)/(x^2(1+1/x^2))dx=int5/(1+1/x^2)dx=#

now set: #1/x^2=t^2# #x=1/t# and #dx=-1/t^2dt#

The integral becomes:

#5int1/(1+t^2)*(-1/t^2)dt=-5[int1/(t^2(1+t^2))dt]=#

#=-5{int[1/t^2-1/(1+t^2)]dt}=#

#=-5[t^(-1)/-1-arctan(t)]=# going back to #x#

#5x+5arctan(1/x)+c#

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Answer 2

Isaiah Allen

To find the antiderivative of (\frac{5x^2}{x^2 + 1}), you can use the method of partial fraction decomposition. First, rewrite the fraction as:

[\frac{5x^2}{x^2 + 1} = \frac{Ax + B}{x^2 + 1}]

Then, multiply both sides by (x^2 + 1) to clear the denominator:

[5x^2 = (Ax + B)(x^2 + 1)]

Expand the right side:

[5x^2 = Ax^3 + Ax + Bx^2 + B]

Now, equate coefficients of like terms:

For (x^2) terms: (0x^2 = B) For (x) terms: (5x = Ax) For constant terms: (0 = A)

From the constant term equation, (A = 0), and from the (x) term equation, (5x = Ax) implies (A = 5).

So, the partial fraction decomposition is:

[\frac{5x^2}{x^2 + 1} = \frac{5}{x^2 + 1}]

Now, you can integrate term by term. The integral of (\frac{5}{x^2 + 1}) is (5\tan^{-1}(x) + C), where (C) is the constant of integration.

Therefore, the antiderivative of (\frac{5x^2}{x^2 + 1}) is (5\tan^{-1}(x) + C).

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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.