What is the integral of 1/(x^2+3x+4) ?

The integral of 1/(x^2+3x+4) is 2/(sqrt(7))arctan((2x+3)/(sqrt(7)))+C

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integral of 1/(x^2+3x+4)

\int \frac{1}{x ^{2} + 3 x + 4} d x

\frac{2}{7} arctan (\frac{2 x + 3}{7}) + C

Solution steps

\int \frac{1}{x ^{2} + 3 x + 4} d x

Complete the square

x^{2} + 3 x + 4 : (x + \frac{3}{2})^{2} + \frac{7}{4}

= \int \frac{1}{( x + \frac{3}{2} ) ^{2} + \frac{7}{4}} d x

Apply u-substitution

: \int \frac{4}{4 u ^{2} + 7} d u

= \int \frac{4}{4 u ^{2} + 7} d u

Take the constant out:

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

= 4 \cdot \int \frac{1}{4 u ^{2} + 7} d u

Apply Integral Substitution

= 4 \cdot \int \frac{1}{2 7 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 4 \cdot \frac{1}{2 7} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 4 \cdot \frac{1}{2 7} arctan (v)

Substitute back

= 4 \cdot \frac{1}{2 7} arctan (\frac{2}{7} (x + \frac{3}{2}))

Simplify

4 \cdot \frac{1}{2 7} arctan (\frac{2}{7} (x + \frac{3}{2})) : \frac{2}{7} arctan (\frac{2 x + 3}{7})

= \frac{2}{7} arctan (\frac{2 x + 3}{7})

Add a constant to the solution

= \frac{2}{7} arctan (\frac{2 x + 3}{7}) + C

\int \frac{1}{81 e ^{- 7 x} + e ^{7 x}} d x

\int \frac{( x ^{2} - 4 )}{x ( x ^{2} + 4 )} d x

\frac{\partial}{\partial x} (y^{2} z)

\int_{0.6}^{1.5} \frac{48 + 8 y}{81} d y

x \to \infty + lim (65.4 - 65.4 e^{(- \frac{7 x}{2180})})

What is the integral of 1/(x^2+3x+4) ?
The integral of 1/(x^2+3x+4) is 2/(sqrt(7))arctan((2x+3)/(sqrt(7)))+C