What is the integral of 5/(4x^2+32x+70) ?

The integral of 5/(4x^2+32x+70) is 5/(2sqrt(6))arctan((sqrt(2))/(sqrt(3))(x+4))+C

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integral of 5/(4x^2+32x+70)

\int \frac{5}{4 x ^{2} + 32 x + 70} d x

\frac{5}{2 6} arctan (\frac{2}{3} (x + 4)) + C

Solution steps

\int \frac{5}{4 x ^{2} + 32 x + 70} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{4 x ^{2} + 32 x + 70} d x

Complete the square

4 x^{2} + 32 x + 70 : 4 (x + 4)^{2} + 6

= 5 \cdot \int \frac{1}{4 ( x + 4 ) ^{2} + 6} d x

Apply u-substitution

: \int \frac{1}{4 u ^{2} + 6} d u

= 5 \cdot \int \frac{1}{4 u ^{2} + 6} d u

Apply Integral Substitution

= 5 \cdot \int \frac{1}{2 6 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 5 \cdot \frac{1}{2 6} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 5 \cdot \frac{1}{2 6} arctan (v)

Substitute back

= 5 \cdot \frac{1}{2 6} arctan (\frac{2}{6} (x + 4))

Simplify

5 \cdot \frac{1}{2 6} arctan (\frac{2}{6} (x + 4)) : \frac{5}{2 6} arctan (\frac{2}{3} (x + 4))

= \frac{5}{2 6} arctan (\frac{2}{3} (x + 4))

Add a constant to the solution

= \frac{5}{2 6} arctan (\frac{2}{3} (x + 4)) + C

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

\int x ln (\frac{( 1 + x )}{( 1 - x )}) d x

\int e^{5 x} 5 d x

\int \frac{x ^{2} + 5}{x} d x

\int \frac{1}{Q} d Q

What is the integral of 5/(4x^2+32x+70) ?
The integral of 5/(4x^2+32x+70) is 5/(2sqrt(6))arctan((sqrt(2))/(sqrt(3))(x+4))+C