What is the integral of (t^3)/(sqrt(a^4+t^4)) ?

The integral of (t^3)/(sqrt(a^4+t^4)) is 1/2 sqrt(a^4+t^4)+C

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integral of (t^3)/(sqrt(a^4+t^4))

\int \frac{t ^{3}}{a ^{4} + t ^{4}} d t

\frac{1}{2} a^{4} + t^{4} + C

Solution steps

\int \frac{t ^{3}}{a ^{4} + t ^{4}} d t

Apply u-substitution

: \int \frac{1}{4 u} d u

= \int \frac{1}{4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= \frac{1}{4} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = a^{4} + t^{4}

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

Simplify

\frac{1}{4} \cdot 2 (a^{4} + t^{4})^{\frac{1}{2}} : \frac{1}{2} a^{4} + t^{4}

= \frac{1}{2} a^{4} + t^{4}

Add a constant to the solution

= \frac{1}{2} a^{4} + t^{4} + C

\frac{d}{d x} (arctan (7 x))

y^{^{'}} = x - y + 1

f (x) = \frac{1300 + 8 x - 0.0004 x ^{2}}{x}

\frac{s + 1}{s ^{2} + 1}

\int (\frac{1 + sin ( x )}{cos ^{2} ( x )}) d x

What is the integral of (t^3)/(sqrt(a^4+t^4)) ?
The integral of (t^3)/(sqrt(a^4+t^4)) is 1/2 sqrt(a^4+t^4)+C