What is the integral of (x^3)/((1+x^4)^{1/2)} ?

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

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integral of (x^3)/((1+x^4)^{1/2)}

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

\frac{x ^{4}}{4 ( 1 + x ^{4} ) ^{\frac{1}{2}}} + \frac{x ^{4} + 2}{4 ( x ^{4} + 1 ) ^{\frac{1}{2}}} + C

Solution steps

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

Apply Integration By Parts

: \frac{x ^{4}}{4 ( 1 + x ^{4} ) ^{\frac{1}{2}}} - \int - \frac{x ^{7}}{2 ( x ^{4} + 1 ) ^{\frac{3}{2}}} d x

= \frac{x ^{4}}{4 ( 1 + x ^{4} ) ^{\frac{1}{2}}} - \int - \frac{x ^{7}}{2 ( x ^{4} + 1 ) ^{\frac{3}{2}}} d x

\int - \frac{x ^{7}}{2 ( x ^{4} + 1 ) ^{\frac{3}{2}}} d x = - \frac{x ^{4} + 2}{4 ( x ^{4} + 1 ) ^{\frac{1}{2}}}

= \frac{x ^{4}}{4 ( 1 + x ^{4} ) ^{\frac{1}{2}}} - (- \frac{x ^{4} + 2}{4 ( x ^{4} + 1 ) ^{\frac{1}{2}}})

Simplify

= \frac{x ^{4}}{4 ( 1 + x ^{4} ) ^{\frac{1}{2}}} + \frac{x ^{4} + 2}{4 ( x ^{4} + 1 ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{x ^{4}}{4 ( 1 + x ^{4} ) ^{\frac{1}{2}}} + \frac{x ^{4} + 2}{4 ( x ^{4} + 1 ) ^{\frac{1}{2}}} + C

f (x) = x^{\frac{5}{2}} e^{x}

\int sec^{2} (x) csc (x) d x

\int 3 sin^{2} (x) cos (x) d x

f (x) = \frac{e ^{x}}{x ^{2}}

\frac{d}{d x} (x - cos (y))

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