What is the integral of (t^3)/(\sqrt[4]{3+t^4)} ?

The integral of (t^3)/(\sqrt[4]{3+t^4)} is 1/3 (3+t^4)^{3/4}+C

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integral of (t^3)/(\sqrt[4]{3+t^4)}

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

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

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

: \frac{4}{3} u^{\frac{3}{4}}

= \frac{1}{4} \cdot \frac{4}{3} u^{\frac{3}{4}}

Substitute back

u = 3 + t^{4}

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

Simplify

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

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

Add a constant to the solution

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

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49 y^{^{''}} - 70 y^{^{'}} + 25 y = 0, y (0) = 2, y^{^{'}} (0) = 3

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What is the integral of (t^3)/(\sqrt[4]{3+t^4)} ?
The integral of (t^3)/(\sqrt[4]{3+t^4)} is 1/3 (3+t^4)^{3/4}+C