What is the integral of 1/(3x^2+4x-7) ?

The integral of 1/(3x^2+4x-7) is -1/10 (ln| 7/5+3/5 x|-ln|-3/5+3/5 x|)+C

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integral of 1/(3x^2+4x-7)

\int \frac{1}{3 x ^{2} + 4 x - 7} d x

- \frac{1}{10} (ln \frac{7}{5} + \frac{3}{5} x - ln - \frac{3}{5} + \frac{3}{5} x) + C

Solution steps

\int \frac{1}{3 x ^{2} + 4 x - 7} d x

Complete the square

3 x^{2} + 4 x - 7 : 3 (x + \frac{2}{3})^{2} - \frac{25}{3}

= \int \frac{1}{3 ( x + \frac{2}{3} ) ^{2} - \frac{25}{3}} d x

Apply u-substitution

: \int \frac{3}{9 u ^{2} - 25} d u

= \int \frac{3}{9 u ^{2} - 25} d u

Take the constant out:

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

= 3 \cdot \int \frac{1}{9 u ^{2} - 25} d u

Apply Integral Substitution

= 3 \cdot \int \frac{1}{15 ( v ^{2} - 1 )} d v

Take the constant out:

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

= 3 \cdot \frac{1}{15} \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

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

= 3 \cdot \frac{1}{15} \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

= 3 \cdot \frac{1}{15} (- \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}

= 3 \cdot \frac{1}{15} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= 3 \cdot \frac{1}{15} (- (\frac{ln \frac{3}{5} ( x + \frac{2}{3} ) + 1}{2} - \frac{ln \frac{3}{5} ( x + \frac{2}{3} ) - 1}{2}))

Simplify

3 \cdot \frac{1}{15} (- (\frac{ln \frac{3}{5} ( x + \frac{2}{3} ) + 1}{2} - \frac{ln \frac{3}{5} ( x + \frac{2}{3} ) - 1}{2})) : - \frac{1}{10} (ln \frac{7}{5} + \frac{3}{5} x - ln - \frac{3}{5} + \frac{3}{5} x)

= - \frac{1}{10} (ln \frac{7}{5} + \frac{3}{5} x - ln - \frac{3}{5} + \frac{3}{5} x)

Add a constant to the solution

= - \frac{1}{10} (ln \frac{7}{5} + \frac{3}{5} x - ln - \frac{3}{5} + \frac{3}{5} x) + C

\frac{d}{d x} (x (x^{5} + 9)^{3})

y = x^{4} - 5 x^{3} + 3 x^{2} - 4 x, (- 1, 13)

\frac{d y}{d x} = \frac{x y + x}{y}

x \to 1000 lim (x^{\frac{2}{3}})

\frac{d}{d x} (\frac{1}{3} x^{3} - 5 x - \frac{4}{x})

What is the integral of 1/(3x^2+4x-7) ?
The integral of 1/(3x^2+4x-7) is -1/10 (ln| 7/5+3/5 x|-ln|-3/5+3/5 x|)+C