What is the integral of ((x+5))/(x^2+10x+26) ?

The integral of ((x+5))/(x^2+10x+26) is 1/2 ln|x^2+10x+26|+C

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integral of ((x+5))/(x^2+10x+26)

\int \frac{( x + 5 )}{x ^{2} + 10 x + 26} d x

\frac{1}{2} ln x^{2} + 10 x + 26 + C

Solution steps

\int \frac{x + 5}{x ^{2} + 10 x + 26} d x

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{1}{2 v} d v

= \int \frac{1}{2 v} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} ln ∣ v ∣

Substitute back

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

Simplify

\frac{1}{2} ln (x + 5)^{2} + 1 : \frac{1}{2} ln x^{2} + 10 x + 26

= \frac{1}{2} ln x^{2} + 10 x + 26

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 10 x + 26 + C

\int \frac{1}{x ^{2} 2 x ^{2} - 32} d x

\int sin^{4} (2 x) \cdot cos (2 x) d x

y = \frac{1}{( 1 + c _{1} e ^{- x} )}

\frac{\partial}{\partial x} (e^{x t})

\frac{d}{d x} (\frac{1}{x ^{\frac{1}{3}}})

What is the integral of ((x+5))/(x^2+10x+26) ?
The integral of ((x+5))/(x^2+10x+26) is 1/2 ln|x^2+10x+26|+C