What is the integral of (10-x)/(x^2+10x+25) ?

The integral of (10-x)/(x^2+10x+25) is -ln|x+5|-(15)/(x+5)+C

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integral of (10-x)/(x^2+10x+25)

\int \frac{10 - x}{x ^{2} + 10 x + 25} d x

- ln ∣ x + 5 ∣ - \frac{15}{x + 5} + C

Solution steps

\int \frac{10 - x}{x ^{2} + 10 x + 25} d x

Complete the square

x^{2} + 10 x + 25 : (x + 5)^{2}

= \int \frac{10 - x}{( x + 5 ) ^{2}} d x

Apply u-substitution

: \int \frac{- u + 15}{u ^{2}} d u

= \int \frac{- u + 15}{u ^{2}} d u

Expand

\frac{- u + 15}{u ^{2}} : - \frac{1}{u} + \frac{15}{u ^{2}}

= \int - \frac{1}{u} + \frac{15}{u ^{2}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{1}{u} d u + \int \frac{15}{u ^{2}} d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{15}{u ^{2}} d u = - \frac{15}{u}

= - ln ∣ u ∣ - \frac{15}{u}

Substitute back

u = x + 5

= - ln ∣ x + 5 ∣ - \frac{15}{x + 5}

Add a constant to the solution

= - ln ∣ x + 5 ∣ - \frac{15}{x + 5} + C

\int \frac{1}{4 x ^{6}} d x

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What is the integral of (10-x)/(x^2+10x+25) ?
The integral of (10-x)/(x^2+10x+25) is -ln|x+5|-(15)/(x+5)+C