What is the integral of 4/(5x^2-3x+10) ?

The integral of 4/(5x^2-3x+10) is 8/(sqrt(191))arctan((10x-3)/(sqrt(191)))+C

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integral of 4/(5x^2-3x+10)

\int \frac{4}{5 x ^{2} - 3 x + 10} d x

\frac{8}{191} arctan (\frac{10 x - 3}{191}) + C

Solution steps

\int \frac{4}{5 x ^{2} - 3 x + 10} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{5 x ^{2} - 3 x + 10} d x

Complete the square

5 x^{2} - 3 x + 10 : 5 (x - \frac{3}{10})^{2} + \frac{191}{20}

= 4 \cdot \int \frac{1}{5 ( x - \frac{3}{10} ) ^{2} + \frac{191}{20}} d x

Apply u-substitution

: \int \frac{20}{100 u ^{2} + 191} d u

= 4 \cdot \int \frac{20}{100 u ^{2} + 191} d u

Take the constant out:

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

= 4 \cdot 20 \cdot \int \frac{1}{100 u ^{2} + 191} d u

Apply Integral Substitution

= 4 \cdot 20 \cdot \int \frac{1}{10 191 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 4 \cdot 20 \cdot \frac{1}{10 191} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 4 \cdot 20 \cdot \frac{1}{10 191} arctan (v)

Substitute back

= 4 \cdot 20 \cdot \frac{1}{10 191} arctan (\frac{10}{191} (x - \frac{3}{10}))

Simplify

4 \cdot 20 \cdot \frac{1}{10 191} arctan (\frac{10}{191} (x - \frac{3}{10})) : \frac{8}{191} arctan (\frac{10 x - 3}{191})

= \frac{8}{191} arctan (\frac{10 x - 3}{191})

Add a constant to the solution

= \frac{8}{191} arctan (\frac{10 x - 3}{191}) + C

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What is the integral of 4/(5x^2-3x+10) ?
The integral of 4/(5x^2-3x+10) is 8/(sqrt(191))arctan((10x-3)/(sqrt(191)))+C