How do you integrate #h(t)=(t+1)/(t^2+2t+2)# using the quotient rule?

Answer 1

Please see the explanation for the steps leading to the answer:

#int(t + 1)/(t^2 + 2t + 2)dx = (1/2)ln(t^2 + 2t + 2) + C#

The quotient rule does not pertain to integrals; it pertains to derivatives.

This integral is best integrated by u substitution:

let #u = t^2 + 2t + 2#, then #du = (2t + 2)dx#, therefore, we shall substitute #(1/2)du# for #(t + 1)dx#:
#int(t + 1)/(t^2 + 2t + 2)dx = (1/2)int(du)/u#
#int(t + 1)/(t^2 + 2t + 2)dx = (1/2)ln|u| + C#

Reverse the substitution for u:

#int(t + 1)/(t^2 + 2t + 2)dx = (1/2)ln|t^2 + 2t + 2| + C#
Because #t^2 + 2t + 2# can never be less than or equal to zero, we may drop the absolute value:
#int(t + 1)/(t^2 + 2t + 2)dx = (1/2)ln(t^2 + 2t + 2) + C#
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Answer 2

To integrate ( h(t) = \frac{t + 1}{t^2 + 2t + 2} ) using the quotient rule, we first express the integrand as a quotient of two functions, then apply the quotient rule for integration.

Step 1: Rewrite the integrand: ( h(t) = \frac{t + 1}{t^2 + 2t + 2} = \frac{t + 1}{(t + 1)^2 + 1} )

Step 2: Apply the quotient rule for integration: Let ( u = t + 1 ) and ( dv = \frac{1}{u^2 + 1} dt ). Then, ( du = dt ) and ( v = \arctan(u) ).

Using the quotient rule for integration: [ \int \frac{u}{u^2 + 1} du = u \arctan(u) - \int \arctan(u) du ]

Step 3: Substitute back the original variable: [ = (t + 1) \arctan(t + 1) - \int \arctan(t + 1) dt ]

Step 4: Integrate the remaining term: [ = (t + 1) \arctan(t + 1) - \int \arctan(t + 1) dt ]

This integral, ( \int \arctan(t + 1) dt ), may not have a closed-form expression in terms of elementary functions. Therefore, this integral cannot be further simplified using elementary functions, and the result is:

[ \int \frac{t + 1}{t^2 + 2t + 2} dt = (t + 1) \arctan(t + 1) - \int \arctan(t + 1) dt + 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.

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.

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