What is the integral of xsqrt(x^4+16) ?

The integral of xsqrt(x^4+16) is 8(1/2 ln((|x^2+sqrt(x^4+16)|)/4)+1/32 x^2sqrt(x^4+16))+C

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integral of xsqrt(x^4+16)

\int x x^{4} + 16 d x

8 (\frac{1}{2} ln (\frac{x ^{2} + x ^{4} + 16}{4}) + \frac{1}{32} x^{2} x^{4} + 16) + C

Solution steps

\int x x^{4} + 16 d x

Simplify

x^{4} : (x^{2})^{2}

= \int x (x^{2})^{2} + 16 d x

Apply u-substitution

: \int \frac{u ^{2} + 16}{2} d u

= \int \frac{u ^{2} + 16}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int u^{2} + 16 d u

Apply Trigonometric Substitution

: \int 16 sec^{3} (v) d v

= \frac{1}{2} \cdot \int 16 sec^{3} (v) d v

Take the constant out:

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

= \frac{1}{2} \cdot 16 \cdot \int sec^{3} (v) d v

Apply Integral Reduction

= \frac{1}{2} \cdot 16 (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} \cdot \int sec (v) d v)

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= \frac{1}{2} \cdot 16 (\frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣)

Substitute back

= \frac{1}{2} \cdot 16 (\frac{sec ^{2} ( arctan ( \frac{1}{4} x ^{2} ) ) sin ( arctan ( \frac{1}{4} x ^{2} ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{4} x^{2})) + sec (arctan (\frac{1}{4} x^{2})))

Simplify

\frac{1}{2} \cdot 16 (\frac{sec ^{2} ( arctan ( \frac{1}{4} x ^{2} ) ) sin ( arctan ( \frac{1}{4} x ^{2} ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{1}{4} x^{2})) + sec (arctan (\frac{1}{4} x^{2}))) : 8 (\frac{1}{2} ln (\frac{x ^{2} + x ^{4} + 16}{4}) + \frac{1}{32} x^{2} x^{4} + 16)

= 8 (\frac{1}{2} ln (\frac{x ^{2} + x ^{4} + 16}{4}) + \frac{1}{32} x^{2} x^{4} + 16)

Add a constant to the solution

= 8 (\frac{1}{2} ln (\frac{x ^{2} + x ^{4} + 16}{4}) + \frac{1}{32} x^{2} x^{4} + 16) + C

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What is the integral of xsqrt(x^4+16) ?
The integral of xsqrt(x^4+16) is 8(1/2 ln((|x^2+sqrt(x^4+16)|)/4)+1/32 x^2sqrt(x^4+16))+C