How do you find the antiderivative of $\int \frac{1}{x^{2} + 10 x + 25} d x$ $int 1/(x^2+10x+25) dx$ ?

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How do you find the antiderivative of $\int \frac{1}{x^{2} + 10 x + 25} d x$ $int 1/(x^2+10x+25) dx$ ?

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Answer 1 · 2016-10-21T21:38:08

$- \frac{1}{x + 5} + C$ $-1/(x+5)+C$

Explanation:

Notice the denominator can be factored:

$\int \frac{1}{x^{2} + 10 x + 25} d x = \int \frac{1}{{(x + 5)}^{2}} d x = \int {(x + 5)}^{- 2} d x$ $int1/(x^2+10x+25)dx=int1/(x+5)^2dx=int(x+5)^-2dx$

Now we can use substitution. Let $u = x + 5$ $u=x+5$ . Thus, $d u = d x$ $du=dx$ . Substituting them in:

$= \int u^{- 2} d u$ $=intu^-2du$

Now use the power rule for integration $\int u^{n} d u = \frac{u^{n + 1}}{n + 1} + C$ $intu^ndu=u^(n+1)/(n+1)+C$

$= \frac{u^{- 2 + 1}}{- 2 + 1} + C = \frac{u^{- 1}}{- 1} + C = - \frac{1}{u} + C = - \frac{1}{x + 5} + C$ $=u^(-2+1)/(-2+1)+C=u^(-1)/(-1)+C=-1/u+C=-1/(x+5)+C$

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How do you find the antiderivative of $\int \frac{1}{x^{2} + 10 x + 25} d x$ $int 1/(x^2+10x+25) dx$ ?

1 Answer

Explanation:

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How do you find the antiderivative of ∫1x2+10x+25dxint 1/(x^2+10x+25) dx?

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How do you find the antiderivative of $\int \frac{1}{x^{2} + 10 x + 25} d x$ $int 1/(x^2+10x+25) dx$ ?