What is the integral of 1/(x^2+10x+25) ?

The integral of 1/(x^2+10x+25) is -1/(x+5)+C

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integral of 1/(x^2+10x+25)

\int \frac{1}{x ^{2} + 10 x + 25} d x

- \frac{1}{x + 5} + C

Solution steps

\int \frac{1}{x ^{2} + 10 x + 25} d x

Take the partial fraction of

\frac{1}{x ^{2} + 10 x + 25} : \frac{1}{( x + 5 ) ^{2}}

= \int \frac{1}{( x + 5 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{u ^{2}} d u

= \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{u}

Substitute back

u = x + 5

= - \frac{1}{x + 5}

Add a constant to the solution

= - \frac{1}{x + 5} + C

n \to \infty lim (\frac{1}{n !})

x \to 0 lim (\frac{1}{x ^{2} ( x + 8 )})

y^{^{''}} + 4 y = sin (2 x)

\int_{0}^{8} 3 x d x

\frac{d y}{d x} + 4 y^{2} = 0

What is the integral of 1/(x^2+10x+25) ?
The integral of 1/(x^2+10x+25) is -1/(x+5)+C