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How do you find the integral of #int 1/(3+(x-2)^2#?

Answer 1

Isaiah Allen

# 1/sqrt3*arc tan((x-2)/sqrt3)+C.#

Let, #I=int1/(3+(x-2)^2)dx.#

Knowing that, #int1/(t^2+a^2)dt=1/a*arc tan(t/a)+c,# we take

substn. #(x-2)=t rArr dx=dt.#

#:. I=int1/(t^2+sqrt3^2)dt,#

#=1/sqrt3*arc tan (t/sqrt3),#

# rArr I=1/sqrt3*arc tan((x-2)/sqrt3)+C.#

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Answer 2

Ezekiel Alexander

To find the integral of ( \int \frac{1}{3+(x-2)^2} ), you can use the substitution method. Let ( u = x - 2 ), then ( du = dx ). Substitute ( u ) into the integral:

[ \int \frac{1}{3+u^2} du ]

Now, this integral resembles the form of the arctangent function. Using the standard integral formula, we can write the integral as:

[ \frac{1}{\sqrt{3}} \arctan\left(\frac{u}{\sqrt{3}}\right) + C ]

Substitute back ( u = x - 2 ):

[ \frac{1}{\sqrt{3}} \arctan\left(\frac{x-2}{\sqrt{3}}\right) + C ]

Therefore, the integral of ( \frac{1}{3+(x-2)^2} ) is ( \frac{1}{\sqrt{3}} \arctan\left(\frac{x-2}{\sqrt{3}}\right) + 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.