What is the integral of 1/(x(8x+9)) ?

The integral of 1/(x(8x+9)) is -1/9 (ln| 16/9 x+2|-ln(16/9 |x|))+C

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integral of 1/(x(8x+9))

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

- \frac{1}{9} (ln \frac{16}{9} x + 2 - ln (\frac{16}{9} ∣ x ∣)) + C

Solution steps

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

Expand

x (8 x + 9) : 8 x^{2} + 9 x

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

Complete the square

8 x^{2} + 9 x : 8 (x + \frac{9}{16})^{2} - \frac{81}{32}

= \int \frac{1}{8 ( x + \frac{9}{16} ) ^{2} - \frac{81}{32}} d x

Apply u-substitution

: \int \frac{32}{256 u ^{2} - 81} d u

= \int \frac{32}{256 u ^{2} - 81} d u

Take the constant out:

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

= 32 \cdot \int \frac{1}{256 u ^{2} - 81} d u

Apply Integral Substitution

= 32 \cdot \int \frac{1}{144 ( v ^{2} - 1 )} d v

Take the constant out:

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

= 32 \cdot \frac{1}{144} \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

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

= 32 \cdot \frac{1}{144} \cdot \int \frac{1}{- ( - v ^{2} + 1 )} d v

Take the constant out:

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

= 32 \cdot \frac{1}{144} (- \int \frac{1}{- v ^{2} + 1} d v)

Use the common integral:

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

= 32 \cdot \frac{1}{144} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= 32 \cdot \frac{1}{144} (- (\frac{ln \frac{16}{9} ( x + \frac{9}{16} ) + 1}{2} - \frac{ln \frac{16}{9} ( x + \frac{9}{16} ) - 1}{2}))

Simplify

32 \cdot \frac{1}{144} (- (\frac{ln \frac{16}{9} ( x + \frac{9}{16} ) + 1}{2} - \frac{ln \frac{16}{9} ( x + \frac{9}{16} ) - 1}{2})) : - \frac{1}{9} (ln \frac{16}{9} x + 2 - ln (\frac{16}{9} ∣ x ∣))

= - \frac{1}{9} (ln \frac{16}{9} x + 2 - ln (\frac{16}{9} ∣ x ∣))

Add a constant to the solution

= - \frac{1}{9} (ln \frac{16}{9} x + 2 - ln (\frac{16}{9} ∣ x ∣)) + C

t^{3} \frac{d y}{d t} + 3 t^{2} y = 2 cos (t), y (π) = 0

2 y^{^{'}} + y = 3 x^{2}

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

\frac{d y}{d x} = y^{2} x

(x^{2} + 3)^{2} y = \frac{( 2 x ( y ^{2} + 1 ) ^{2} )}{( x ^{2} + 3 ) y ^{^{'}}} - (x^{4} + 6 x^{2} + 9) y

What is the integral of 1/(x(8x+9)) ?
The integral of 1/(x(8x+9)) is -1/9 (ln| 16/9 x+2|-ln(16/9 |x|))+C