What is the integral of 1/(4a^2+12a+5) ?

The integral of 1/(4a^2+12a+5) is -1/8 (ln|a+5/2 |-ln|a+1/2 |)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(4a^2+12a+5)

\int \frac{1}{4 a ^{2} + 12 a + 5} d a

- \frac{1}{8} (ln a + \frac{5}{2} - ln a + \frac{1}{2}) + C

Solution steps

\int \frac{1}{4 a ^{2} + 12 a + 5} d a

Complete the square

4 a^{2} + 12 a + 5 : 4 (a + \frac{3}{2})^{2} - 4

= \int \frac{1}{4 ( a + \frac{3}{2} ) ^{2} - 4} d a

Apply u-substitution

: \int \frac{1}{4 u ^{2} - 4} d u

= \int \frac{1}{4 u ^{2} - 4} d u

\frac{1}{4 u ^{2} - 4} = \frac{1}{4} \frac{1}{u ^{2} - 1}

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

Take the constant out:

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

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

Factor

u^{2} - 1 : - (- u^{2} + 1)

= \frac{1}{4} \cdot \int \frac{1}{- ( - u ^{2} + 1 )} d u

Take the constant out:

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

= \frac{1}{4} (- \int \frac{1}{- u ^{2} + 1} d u)

Use the common integral:

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

= \frac{1}{4} (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = a + \frac{3}{2}

= \frac{1}{4} (- (\frac{ln a + \frac{3}{2} + 1}{2} - \frac{ln a + \frac{3}{2} - 1}{2}))

Simplify

\frac{1}{4} (- (\frac{ln a + \frac{3}{2} + 1}{2} - \frac{ln a + \frac{3}{2} - 1}{2})) : - \frac{1}{8} (ln a + \frac{5}{2} - ln a + \frac{1}{2})

= - \frac{1}{8} (ln a + \frac{5}{2} - ln a + \frac{1}{2})

Add a constant to the solution

= - \frac{1}{8} (ln a + \frac{5}{2} - ln a + \frac{1}{2}) + C

\int cos^{3} (5 x) \cdot sin^{2} (5 x) d x

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

\int \frac{( 2 x + 8 )}{( x ^{2} + 2 x + 5 )} d x

\int \frac{1}{x ^{3}} csc^{4} (\frac{1}{x ^{2}}) d x

\int \frac{6}{5} (2 - x) d x

What is the integral of 1/(4a^2+12a+5) ?
The integral of 1/(4a^2+12a+5) is -1/8 (ln|a+5/2 |-ln|a+1/2 |)+C