What is the integral from 1/12 to 1/4 of 2/(sqrt(x)+4x\sqrt{x)} ?

The integral from 1/12 to 1/4 of 2/(sqrt(x)+4x\sqrt{x)} is pi/6

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integral from 1/12 to 1/4 of 2/(sqrt(x)+4x\sqrt{x)}

\int_{\frac{1}{12}}^{\frac{1}{4}} \frac{2}{x + 4 x x} d x

\frac{π}{6}

Solution steps

\int_{\frac{1}{12}}^{\frac{1}{4}} \frac{2}{x + 4 x x} d x

Take the constant out:

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

= 2 \cdot \int_{\frac{1}{12}}^{\frac{1}{4}} \frac{1}{x + 4 x x} d x

\frac{1}{x + 4 x x} = \frac{x - 4 x x}{( x ) ^{2} - ( 4 x x ) ^{2}}

= 2 \cdot \int_{\frac{1}{12}}^{\frac{1}{4}} \frac{x - 4 x x}{( x ) ^{2} - ( 4 x x ) ^{2}} d x

Simplify

\frac{x - 4 x x}{( x ) ^{2} - ( 4 x x ) ^{2}} : \frac{1}{x ( 4 x + 1 )}

= 2 \cdot \int_{\frac{1}{12}}^{\frac{1}{4}} \frac{1}{x ( 4 x + 1 )} d x

Apply u-substitution

: \int_{\frac{1}{2 3}}^{\frac{1}{2}} \frac{2}{4 u ^{2} + 1} d u

= 2 \cdot \int_{\frac{1}{2 3}}^{\frac{1}{2}} \frac{2}{4 u ^{2} + 1} d u

Take the constant out:

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

= 2 \cdot 2 \cdot \int_{\frac{1}{2 3}}^{\frac{1}{2}} \frac{1}{4 u ^{2} + 1} d u

Apply Integral Substitution

= 2 \cdot 2 \cdot \int_{\frac{1}{3}}^{1} \frac{1}{2 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 2 \cdot 2 \cdot \frac{1}{2} \cdot \int_{\frac{1}{3}}^{1} \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 2 \cdot 2 \cdot \frac{1}{2} [arctan (v)]_{\frac{1}{3}}^{1}

Simplify

2 \cdot 2 \cdot \frac{1}{2} [arctan (v)]_{\frac{1}{3}}^{1} : 2 [arctan (v)]_{\frac{1}{3}}^{1}

= 2 [arctan (v)]_{\frac{1}{3}}^{1}

Compute the boundaries:

\frac{π}{12}

= 2 \cdot \frac{π}{12}

Simplify

= \frac{π}{6}

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What is the integral from 1/12 to 1/4 of 2/(sqrt(x)+4x\sqrt{x)} ?
The integral from 1/12 to 1/4 of 2/(sqrt(x)+4x\sqrt{x)} is pi/6