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

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Answer 1 · 2018-03-19T17:16:28

# 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#.