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Popular >

integral of 1/(sqrt(12x+0.02x^2))

Solution

\int \frac{1}{12 x + 0.02 x ^{2}} d x

Solution

7.07106 \dots ln 0.00001 \dots x^{2} + 0.00666 \dots x - 2.0 E - 16 + \frac{1}{6} (0.02 x + 6) + C

Solution steps

\int \frac{1}{12 x + 0.02 x ^{2}} d x

Factor

\frac{1}{12 x + 0.02 x ^{2}}

= \int \frac{1}{x ( 0.02 x + 12 )} d x

Apply u-substitution

: \int \frac{1}{2 \cdot 0.1 u ( u - 12 )} d u

= \int \frac{1}{2 \cdot 0.1 u ( u - 12 )} d u

Take the constant out:

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

= \frac{1}{2 \cdot 0.1} \cdot \int \frac{1}{u ( u - 12 )} d u

Simplify

= 7.07106 \dots \cdot \int \frac{1}{u ( u - 12 )} d u

Complete the square

u (u - 12) : (u - 6)^{2} - 36

= 7.07106 \dots \cdot \int \frac{1}{( u - 6 ) ^{2} - 36} d u

Apply u-substitution

: \int \frac{1}{v ^{2} - 36} d v

= 7.07106 \dots \cdot \int \frac{1}{v ^{2} - 36} d v

Apply Trigonometric Substitution

: \int sec (w) d w

= 7.07106 \dots \cdot \int sec (w) d w

Use the common integral:

\int sec (w) d w = ln ∣ tan (w) + sec (w) ∣

= 7.07106 \dots ln ∣ tan (w) + sec (w) ∣

Substitute back

= 7.07106 \dots ln tan (arcsec (\frac{1}{6} (0.02 x + 12 - 6))) + sec (arcsec (\frac{1}{6} (0.02 x + 12 - 6)))

Simplify

7.07106 \dots ln tan (arcsec (\frac{1}{6} (0.02 x + 12 - 6))) + sec (arcsec (\frac{1}{6} (0.02 x + 12 - 6))) : 7.07106 \dots ln 0.00001 \dots x^{2} + 0.00666 \dots x - 2.0 E - 16 + \frac{1}{6} (0.02 x + 6)

= 7.07106 \dots ln 0.00001 \dots x^{2} + 0.00666 \dots x - 2.0 E - 16 + \frac{1}{6} (0.02 x + 6)

Add a constant to the solution

= 7.07106 \dots ln 0.00001 \dots x^{2} + 0.00666 \dots x - 2.0 E - 16 + \frac{1}{6} (0.02 x + 6) + C

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