What is the integral of 1/(xsqrt(-ln^2(x)+ln(x)+3)) ?

The integral of 1/(xsqrt(-ln^2(x)+ln(x)+3)) is arcsin(1/(sqrt(13))(2ln(x)-1))+C

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integral of 1/(xsqrt(-ln^2(x)+ln(x)+3))

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

arcsin (\frac{1}{13} (2 ln (x) - 1)) + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{- u ^{2} + u + 3} d u

= \int \frac{1}{- u ^{2} + u + 3} d u

Complete the square

- u^{2} + u + 3 : - (u - \frac{1}{2})^{2} + \frac{13}{4}

= \int \frac{1}{- ( u - \frac{1}{2} ) ^{2} + \frac{13}{4}} d u

Apply u-substitution

: \int \frac{2}{- 4 v ^{2} + 13} d v

= \int \frac{2}{- 4 v ^{2} + 13} d v

Take the constant out:

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

= 2 \cdot \int \frac{1}{- 4 v ^{2} + 13} d v

Apply Trigonometric Substitution

: \int \frac{1}{2} d w

= 2 \cdot \int \frac{1}{2} d w

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int 1 d w

Integral of a constant:

\int a d x = a x

= 2 \cdot \frac{1}{2} \cdot 1 \cdot w

Substitute back

= 2 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{2}{13} (ln (x) - \frac{1}{2}))

Simplify

2 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{2}{13} (ln (x) - \frac{1}{2})) : arcsin (\frac{1}{13} (2 ln (x) - 1))

= arcsin (\frac{1}{13} (2 ln (x) - 1))

Add a constant to the solution

= arcsin (\frac{1}{13} (2 ln (x) - 1)) + C

\frac{d}{d t} (2 e^{t} + 2 e^{2 t})

\int \frac{2 x + 5}{9 - x ^{2}} d x

y = sin (e^{2 x})

\frac{d}{d x} (\frac{x + 3}{8 x ^{2} e ^{x}})

x \to 0 + lim (2^{x})

What is the integral of 1/(xsqrt(-ln^2(x)+ln(x)+3)) ?
The integral of 1/(xsqrt(-ln^2(x)+ln(x)+3)) is arcsin(1/(sqrt(13))(2ln(x)-1))+C