What is the integral of (8(x^2+1))/(sqrt(4x^2+9)) ?

The integral of (8(x^2+1))/(sqrt(4x^2+9)) is xsqrt(4x^2+9)-1/2 ln(1/3 |2x+sqrt(4x^2+9)|)+C

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integral of (8(x^2+1))/(sqrt(4x^2+9))

\int \frac{8 ( x ^{2} + 1 )}{4 x ^{2} + 9} d x

x 4 x^{2} + 9 - \frac{1}{2} ln (\frac{1}{3} 2 x + 4 x^{2} + 9) + C

Solution steps

\int \frac{8 ( x ^{2} + 1 )}{4 x ^{2} + 9} d x

Take the constant out:

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

= 8 \cdot \int \frac{x ^{2} + 1}{4 x ^{2} + 9} d x

Expand

\frac{x ^{2} + 1}{4 x ^{2} + 9} : \frac{x ^{2}}{4 x ^{2} + 9} + \frac{1}{4 x ^{2} + 9}

= 8 \cdot \int \frac{x ^{2}}{4 x ^{2} + 9} + \frac{1}{4 x ^{2} + 9} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 8 (\int \frac{x ^{2}}{4 x ^{2} + 9} d x + \int \frac{1}{4 x ^{2} + 9} d x)

\int \frac{x ^{2}}{4 x ^{2} + 9} d x = \frac{1}{16} (- 9 ln (\frac{2 x + 4 x ^{2} + 9}{3}) + 2 x 4 x^{2} + 9)

\int \frac{1}{4 x ^{2} + 9} d x = \frac{1}{2} ln (\frac{2 x + 9 + 4 x ^{2}}{3})

= 8 (\frac{1}{16} (- 9 ln (\frac{2 x + 4 x ^{2} + 9}{3}) + 2 x 4 x^{2} + 9) + \frac{1}{2} ln (\frac{2 x + 9 + 4 x ^{2}}{3}))

Simplify

8 (\frac{1}{16} (- 9 ln (\frac{2 x + 4 x ^{2} + 9}{3}) + 2 x 4 x^{2} + 9) + \frac{1}{2} ln (\frac{2 x + 9 + 4 x ^{2}}{3})) : x 4 x^{2} + 9 - \frac{1}{2} ln (\frac{1}{3} 2 x + 4 x^{2} + 9)

= x 4 x^{2} + 9 - \frac{1}{2} ln (\frac{1}{3} 2 x + 4 x^{2} + 9)

Add a constant to the solution

= x 4 x^{2} + 9 - \frac{1}{2} ln (\frac{1}{3} 2 x + 4 x^{2} + 9) + C

\frac{1 + 6 s ^{2}}{s ( s ^{2} + 1 )}

\frac{d ^{2} y}{d x ^{2}} = (4 sin (3 x))

\frac{s}{( s ^{2} + 4 ) ( s + 2 )}

\int - 4 x - 1 d x

y = sin (x^{5})

What is the integral of (8(x^2+1))/(sqrt(4x^2+9)) ?
The integral of (8(x^2+1))/(sqrt(4x^2+9)) is xsqrt(4x^2+9)-1/2 ln(1/3 |2x+sqrt(4x^2+9)|)+C