What is the integral of (-0.33x)/(sqrt(0.02x^2+10)) ?

The integral of (-0.33x)/(sqrt(0.02x^2+10)) is -52.17758…sqrt(0.002x^2+1)+C

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integral of (-0.33x)/(sqrt(0.02x^2+10))

\int \frac{- 0.33 x}{0.02 x ^{2} + 10} d x

- 52.17758 \dots 0.002 x^{2} + 1 + C

Solution steps

\int \frac{- 0.33 x}{0.02 x ^{2} + 10} d x

Take the constant out:

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

= - 0.33 \cdot \int \frac{x}{0.02 x ^{2} + 10} d x

0.02 x^{2} + 10 = 10 (\frac{0.02 x ^{2}}{10} + 1)

= - 0.33 \cdot \int \frac{x}{10 ( \frac{0.02 x ^{2}}{10} + 1 )} d x

Apply exponent rule:

\frac{a ^{n}}{b ^{n}} = (\frac{a}{b})^{n}

\frac{0.02 x ^{2}}{10} = (0.04472 \dots x)^{2}

= - 0.33 \cdot \int \frac{x}{10 ( ( 0.04472 \dots x ) ^{2} + 1 )} d x

Apply exponent rule:

(a \cdot b)^{n} = a^{n} \cdot b^{n}

10 ((0.04472 \dots x)^{2} + 1) = 10 ((0.04472 \dots x)^{2} + 1)

= - 0.33 \cdot \int \frac{x}{10 ( 0.04472 \dots x ) ^{2} + 1} d x

Take the constant out:

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

= - 0.33 \cdot \frac{1}{10} \cdot \int \frac{x}{( 0.04472 \dots x ) ^{2} + 1} d x

(0.04472 \dots x)^{2} = 0.002 x^{2}

= - 0.33 \cdot \frac{1}{10} \cdot \int \frac{x}{0.002 x ^{2} + 1} d x

Apply u-substitution

: \int \frac{1}{0.004 u} d u

= - 0.33 \cdot \frac{1}{10} \cdot \int \frac{1}{0.004 u} d u

Take the constant out:

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

= - 0.33 \cdot \frac{1}{10} \cdot \frac{1}{0.004} \cdot \int \frac{1}{u} d u

Simplify

= - 0.33 \cdot \frac{1}{10} \cdot 250 \cdot \int \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= - 0.33 \cdot \frac{1}{10} \cdot 250 \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 0.002 x^{2} + 1

= - 0.33 \cdot \frac{1}{10} \cdot 250 \cdot 2 (0.002 x^{2} + 1)^{\frac{1}{2}}

Simplify

- 0.33 \cdot \frac{1}{10} \cdot 250 \cdot 2 (0.002 x^{2} + 1)^{\frac{1}{2}} : - 52.17758 \dots 0.002 x^{2} + 1

= - 52.17758 \dots 0.002 x^{2} + 1

Add a constant to the solution

= - 52.17758 \dots 0.002 x^{2} + 1 + C

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What is the integral of (-0.33x)/(sqrt(0.02x^2+10)) ?
The integral of (-0.33x)/(sqrt(0.02x^2+10)) is -52.17758…sqrt(0.002x^2+1)+C