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

1 Answer
Mar 19, 2018

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

Explanation:

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.