What is the integral of 7/(sqrt(16x^2+24x+9)) ?

The integral of 7/(sqrt(16x^2+24x+9)) is 7/4 ln|x+3/4 |+C

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integral of 7/(sqrt(16x^2+24x+9))

\int \frac{7}{16 x ^{2} + 24 x + 9} d x

\frac{7}{4} ln x + \frac{3}{4} + C

Solution steps

\int \frac{7}{16 x ^{2} + 24 x + 9} d x

Take the constant out:

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

= 7 \cdot \int \frac{1}{16 x ^{2} + 24 x + 9} d x

Complete the square

16 x^{2} + 24 x + 9 : 16 (x + \frac{3}{4})^{2}

= 7 \cdot \int \frac{1}{16 ( x + \frac{3}{4} ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{4 u} d u

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 7 \cdot \frac{1}{4} ln ∣ u ∣

Substitute back

u = x + \frac{3}{4}

= 7 \cdot \frac{1}{4} ln x + \frac{3}{4}

Simplify

7 \cdot \frac{1}{4} ln x + \frac{3}{4} : \frac{7}{4} ln x + \frac{3}{4}

= \frac{7}{4} ln x + \frac{3}{4}

Add a constant to the solution

= \frac{7}{4} ln x + \frac{3}{4} + C

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What is the integral of 7/(sqrt(16x^2+24x+9)) ?
The integral of 7/(sqrt(16x^2+24x+9)) is 7/4 ln|x+3/4 |+C