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

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

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integral of sqrt(4+25x^2)

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

y = x^{3}, (1, 0)

\frac{d}{d x} (sin (3 x + y z))

\frac{d}{d x} (x^{0.7})

\int_{0}^{2 π} t^{2} sin (2) d t

\frac{\partial}{\partial v} (sin (u) sin (v))

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