What is the integral of sqrt(1+25x^2) ?

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

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integral of sqrt(1+25x^2)

\int 1 + 25 x^{2} d x

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

Solution steps

\int 1 + 25 x^{2} d x

Apply Trigonometric Substitution

: \int \frac{1}{5} sec^{3} (u) d u

= \int \frac{1}{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{1}{5} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

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

Substitute back

u = arctan (5 x)

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (x^{2} sin (6 x))

\frac{d}{d x} (ln (x^{2} + x))

\frac{d}{d x} (cos (x) \cdot sin (x))

y^{^{''}} + 9 y = x e^{2 x}

\frac{\partial}{\partial x} (\frac{3}{( x ^{2} + z ^{2} )})

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