What is the integral of tsqrt(1+4t^2) ?

The integral of tsqrt(1+4t^2) is 1/12 (1+4t^2)^{3/2}+C

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integral of tsqrt(1+4t^2)

\int t 1 + 4 t^{2} d t

\frac{1}{12} (1 + 4 t^{2})^{\frac{3}{2}} + C

Solution steps

\int t 1 + 4 t^{2} d t

Apply u-substitution

: \int \frac{u}{8} d u

= \int \frac{u}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{8} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 1 + 4 t^{2}

= \frac{1}{8} \cdot \frac{2}{3} (1 + 4 t^{2})^{\frac{3}{2}}

Simplify

\frac{1}{8} \cdot \frac{2}{3} (1 + 4 t^{2})^{\frac{3}{2}} : \frac{1}{12} (1 + 4 t^{2})^{\frac{3}{2}}

= \frac{1}{12} (1 + 4 t^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{12} (1 + 4 t^{2})^{\frac{3}{2}} + C

y^{^{'}} = \frac{( 1 + x )}{x y ^{4}}, y (1) = 3

a (1 + e^{(b x + c)}) + d

\frac{d}{d x} (- 2 x - y^{2} cos (x y^{2}) - 1)

\frac{1}{5} x^{5} - \frac{1}{2 x ^{4}} + 3 x + 9

\int \frac{4 x + 1}{4 x ^{2} + 2 x + 3} d x

What is the integral of tsqrt(1+4t^2) ?
The integral of tsqrt(1+4t^2) is 1/12 (1+4t^2)^{3/2}+C