What is the integral of x/(sqrt(49+x^4)) ?

The integral of x/(sqrt(49+x^4)) is 1/2 ln((|x^2+sqrt(49+x^4)|)/7)+C

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integral of x/(sqrt(49+x^4))

\int \frac{x}{49 + x ^{4}} d x

\frac{1}{2} ln (\frac{x ^{2} + 49 + x ^{4}}{7}) + C

Solution steps

\int \frac{x}{49 + x ^{4}} d x

Apply u-substitution

: \int \frac{1}{2 49 + u ^{2}} d u

= \int \frac{1}{2 49 + u ^{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 \frac{1}{49 + u ^{2}} d u

Apply Trigonometric Substitution

: \int sec (v) d v

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

Use the common integral:

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

= \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

Substitute back

= \frac{1}{2} ln tan (arctan (\frac{1}{7} x^{2})) + sec (arctan (\frac{1}{7} x^{2}))

Simplify

\frac{1}{2} ln tan (arctan (\frac{1}{7} x^{2})) + sec (arctan (\frac{1}{7} x^{2})) : \frac{1}{2} ln (\frac{x ^{2} + 49 + x ^{4}}{7})

= \frac{1}{2} ln (\frac{x ^{2} + 49 + x ^{4}}{7})

Add a constant to the solution

= \frac{1}{2} ln (\frac{x ^{2} + 49 + x ^{4}}{7}) + C

\frac{2}{t} - \frac{4}{t ^{2}}

\int 2 x^{2} - 4 x d x

\int_{0}^{5} 40 e^{- 0.3 t} d t

y^{^{'}} = 7 x e^{y}, y (0) = 0

x \to 0 lim (\frac{a ^{x} - 1}{2 x})

What is the integral of x/(sqrt(49+x^4)) ?
The integral of x/(sqrt(49+x^4)) is 1/2 ln((|x^2+sqrt(49+x^4)|)/7)+C