What is the integral of 1/(x^2sqrt(1+4x^2)) ?

The integral of 1/(x^2sqrt(1+4x^2)) is -(sqrt(1+4x^2))/x+C

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integral of 1/(x^2sqrt(1+4x^2))

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

- \frac{1 + 4 x ^{2}}{x} + C

Solution steps

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

Apply Trigonometric Substitution

: \int 2 cot^{2} (u) sec (u) d u

= \int 2 cot^{2} (u) sec (u) d u

Take the constant out:

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

= 2 \cdot \int cot^{2} (u) sec (u) d u

Simplify

cot^{2} (u) sec (u) : cot (u) csc (u)

= 2 \cdot \int cot (u) csc (u) d u

Apply u-substitution

: \int - 1 d v

= 2 \cdot \int - 1 d v

Integral of a constant:

\int a d x = a x

= 2 (- 1) v

Substitute back

= 2 (- 1) csc (arctan (2 x))

Simplify

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

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

Add a constant to the solution

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

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What is the integral of 1/(x^2sqrt(1+4x^2)) ?
The integral of 1/(x^2sqrt(1+4x^2)) is -(sqrt(1+4x^2))/x+C