What is the integral of 1/(4x^2+48x+148) ?

The integral of 1/(4x^2+48x+148) is 1/4 arctan(x+6)+C

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integral of 1/(4x^2+48x+148)

\int \frac{1}{4 x ^{2} + 48 x + 148} d x

\frac{1}{4} arctan (x + 6) + C

Solution steps

\int \frac{1}{4 x ^{2} + 48 x + 148} d x

Complete the square

4 x^{2} + 48 x + 148 : 4 (x + 6)^{2} + 4

= \int \frac{1}{4 ( x + 6 ) ^{2} + 4} d x

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

Use the common integral:

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

= \frac{1}{4} arctan (u)

Substitute back

u = x + 6

= \frac{1}{4} arctan (x + 6)

Add a constant to the solution

= \frac{1}{4} arctan (x + 6) + C

\frac{\partial}{\partial x} (e^{(z + e^{x} + 1)})

\frac{\partial}{\partial x} (- 3 cos (3 x))

\int_{6}^{8} \frac{80}{( x - 6 ) ^{3}} d x

\frac{\partial}{\partial y} (x^{2} - y^{3} + 5 x + 12 y + 1)

2 x^{2} + 3

What is the integral of 1/(4x^2+48x+148) ?
The integral of 1/(4x^2+48x+148) is 1/4 arctan(x+6)+C