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What is #int (5x+16)/(x^2 +6x +34) dx#?

Answer 1

Charles Arocha

# 5/2ln|(x^2+6x+34)|+1/5arc tan((x+3)/5)+C#.

Suppose that, #I=int(5x+16)/(x^2+6x+34)dx#.

In such cases, we have to find #m,n in RR#, such that,

#5x+16=md/dx(x^2+6x+34)+n#, .

#:. 5x+16=m(2x+6)+n=(2m)x+(6m+n)#.

Comparing the respective coefficients, we get,

#2m=5 [rArr m=5/2] and 6m+n=16#.

#:. 6(5/2)+n=16 rArr n=1#.

Replacing #5x+16# by #{5/2d/dx(x^2+6x+34)+1}#, we have,

# I=int{5/2d/dx(x^2+6x+34)+1}/(x^2+6x+34)dx#,

#=5/2int{d/dx(x^2+6x+34)}/(x^2+6x+34)dx+int1/(x^2+6x+34)dx#,

#=5/2ln|(x^2+6x+34)|+int1/{(x+3)^2+5^2}dx#,

#=5/2ln|(x^2+6x+34)|+1/5arc tan((x+3)/5)#.

#rArr I=5/2ln|(x^2+6x+34)|+1/5arc tan((x+3)/5)+C#.

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Answer 2

Axel Alvarez

To integrate (\frac{5x+16}{x^2 +6x +34}) with respect to (x), we first need to perform partial fraction decomposition to break down the fraction into simpler components. Then, we can integrate each component separately. The integral involves arctangent and natural logarithm functions. The final result is:

[ \int \frac{5x + 16}{x^2 + 6x + 34} , dx = \frac{5}{2} \ln|x^2 + 6x + 34| + \frac{16}{\sqrt{125}} \arctan{\left(\frac{x+3}{\sqrt{125}}\right)} + 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.