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

The integral of 1/(2x^2+8x+25) is 1/(sqrt(34))arctan(sqrt(2/17)(x+2))+C

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

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

\frac{1}{34} arctan (\frac{2}{17} (x + 2)) + C

Solution steps

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

Complete the square

2 x^{2} + 8 x + 25 : 2 (x + 2)^{2} + 17

= \int \frac{1}{2 ( x + 2 ) ^{2} + 17} d x

Apply u-substitution

: \int \frac{1}{2 u ^{2} + 17} d u

= \int \frac{1}{2 u ^{2} + 17} d u

Apply Integral Substitution

= \int \frac{1}{34 ( 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}{34} \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}{34} arctan (v)

Substitute back

= \frac{1}{34} arctan (\frac{2}{17} (x + 2))

Combine same powers :

\frac{x}{y} = \frac{x}{y}

= \frac{1}{34} arctan (\frac{2}{17} (x + 2))

Add a constant to the solution

= \frac{1}{34} arctan (\frac{2}{17} (x + 2)) + C

\frac{d y}{d x} = - \frac{5 x}{y}

\frac{d}{d x} (e^{t a n (2 x + 1)})

x \to 5 - lim (∣ x - 5 ∣)

\int - cos (x) + sin (x) + 2 d x

\frac{d}{d x} (- 5 x + 5)

What is the integral of 1/(2x^2+8x+25) ?
The integral of 1/(2x^2+8x+25) is 1/(sqrt(34))arctan(sqrt(2/17)(x+2))+C