What is the integral of sqrt(25x^2+4) ?

The integral of sqrt(25x^2+4) is 1/10 (5xsqrt(25x^2+4)+4ln((|5x+sqrt(25x^2+4)|)/2))+C

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integral of sqrt(25x^2+4)

\int 25 x^{2} + 4 d x

\frac{1}{10} (5 x 25 x^{2} + 4 + 4 ln (\frac{5 x + 25 x ^{2} + 4}{2})) + C

Solution steps

\int 25 x^{2} + 4 d x

Apply Trigonometric Substitution

: \int \frac{4}{5} sec^{3} (u) d u

= \int \frac{4}{5} sec^{3} (u) d u

Take the constant out:

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

= \frac{4}{5} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= \frac{4}{5} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} \cdot \int sec (u) d u)

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

= \frac{4}{5} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (\frac{5}{2} x)

= \frac{4}{5} (\frac{sec ^{2} ( arctan ( \frac{5}{2} x ) ) sin ( arctan ( \frac{5}{2} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{5}{2} x)) + sec (arctan (\frac{5}{2} x)))

Simplify

\frac{4}{5} (\frac{sec ^{2} ( arctan ( \frac{5}{2} x ) ) sin ( arctan ( \frac{5}{2} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{5}{2} x)) + sec (arctan (\frac{5}{2} x))) : \frac{1}{10} (5 x 25 x^{2} + 4 + 4 ln (\frac{5 x + 25 x ^{2} + 4}{2}))

= \frac{1}{10} (5 x 25 x^{2} + 4 + 4 ln (\frac{5 x + 25 x ^{2} + 4}{2}))

Add a constant to the solution

= \frac{1}{10} (5 x 25 x^{2} + 4 + 4 ln (\frac{5 x + 25 x ^{2} + 4}{2})) + C

\int_{- l n (4)}^{0} 3 e^{x} d x

2 cos (x) + cos^{2} (x)

\int 9 x^{2} + 1 d x

\int \frac{1}{x ^{2} - 4 x - 12} d x

\frac{20}{s ( s ^{2} + 4 )}

What is the integral of sqrt(25x^2+4) ?
The integral of sqrt(25x^2+4) is 1/10 (5xsqrt(25x^2+4)+4ln((|5x+sqrt(25x^2+4)|)/2))+C