What is the integral of 1/(x^2+8x+25) ?

The integral of 1/(x^2+8x+25) is 1/3 arctan((x+4)/3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(x^2+8x+25)

\int \frac{1}{x ^{2} + 8 x + 25} d x

\frac{1}{3} arctan (\frac{x + 4}{3}) + C

Solution steps

\int \frac{1}{x ^{2} + 8 x + 25} d x

Complete the square

x^{2} + 8 x + 25 : (x + 4)^{2} + 9

= \int \frac{1}{( x + 4 ) ^{2} + 9} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 9} d u

= \int \frac{1}{u ^{2} + 9} d u

Apply Integral Substitution

= \int \frac{1}{3 ( v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= \frac{1}{3} arctan (v)

Substitute back

= \frac{1}{3} arctan (\frac{x + 4}{3})

Add a constant to the solution

= \frac{1}{3} arctan (\frac{x + 4}{3}) + C

x y^{^{'}} + (1 - x) y = 2 x e^{x}

x \to 4 - lim (\frac{5}{x - 4})

\int \frac{2 x}{1 - 4 x ^{2}} d x

\frac{\partial}{\partial y} (x^{3} y - 3 x y^{2})

\int_{0}^{7} \frac{1}{49 + t ^{2}} d t

What is the integral of 1/(x^2+8x+25) ?
The integral of 1/(x^2+8x+25) is 1/3 arctan((x+4)/3)+C