How do you find the antiderivative of $\int \frac{1}{x^{2} + 10 x + 21} d x$ $int 1/(x^2+10x+21) dx$ ?

Question

How do you find the antiderivative of $\int \frac{1}{x^{2} + 10 x + 21} d x$ $int 1/(x^2+10x+21) dx$ ?

1 Answer

Impact of this question

2193 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-12T20:01:09

$\int \frac{1}{x^{2} + 10 x + 21} d x = \frac{1}{4} ln | x + 3 | + \frac{1}{4} ln | x + 7 | + C$ $int 1/(x^2+10x+21) dx = 1/4 ln abs(x+3) + 1/4 ln abs(x+7) + C$

Explanation:

$\frac{1}{x^{2} + 10 x + 21} = \frac{1}{(x + 3) (x + 7)}$ $1/(x^2+10x+21) = 1/((x+3)(x+7))$

$\frac{1}{x^{2} + 10 x + 21} = \frac{1}{4} (\frac{(x + 7) - (x + 3)}{(x + 3) (x + 7)})$ $color(white)(1/(x^2+10x+21)) = 1/4(((x+7)-(x+3))/((x+3)(x+7)))$

$\frac{1}{x^{2} + 10 x + 21} = \frac{1}{4} (\frac{1}{x + 3} - \frac{1}{x + 7})$ $color(white)(1/(x^2+10x+21)) = 1/4(1/(x+3)-1/(x+7))$

So:

$\int \frac{1}{x^{2} + 10 x + 21} d x = \int \frac{1}{4} (\frac{1}{x + 3} - \frac{1}{x + 7}) d x$ $int 1/(x^2+10x+21) dx = int 1/4(1/(x+3)-1/(x+7)) dx$

$\int \frac{1}{x^{2} + 10 x + 21} d x = \frac{1}{4} ln | x + 3 | + \frac{1}{4} ln | x + 7 | + C$ $color(white)(int 1/(x^2+10x+21) dx) = 1/4 ln abs(x+3) + 1/4 ln abs(x+7) + C$

Science

Math

Humanities

... and beyond

How do you find the antiderivative of $\int \frac{1}{x^{2} + 10 x + 21} d x$ $int 1/(x^2+10x+21) dx$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the antiderivative of ∫1x2+10x+21dxint 1/(x^2+10x+21) dx?

1 Answer

Explanation:

Related questions

Impact of this question

How do you find the antiderivative of $\int \frac{1}{x^{2} + 10 x + 21} d x$ $int 1/(x^2+10x+21) dx$ ?