What is the integral of 7x*4^{sqrt(x)} ?

The integral of 7x*4^{sqrt(x)} is

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integral of 7x*4^{sqrt(x)}

\int 7 x \cdot 4^{x} d x

\frac{7}{8 ln ^{4} ( 2 )} (ln^{3} (2) \cdot 2^{2 x + 3} x^{\frac{3}{2}} - 3 (ln^{2} (2) \cdot 4^{x + 1} x - 2 (ln (2) \cdot 2^{2 x + 1} x - 4^{x}))) + C

Solution steps

\int 7 x \cdot 4^{x} d x

Take the constant out:

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

= 7 \cdot \int x \cdot 4^{x} d x

Apply u-substitution

: \int \frac{ln ^{3} ( u )}{8 ln ^{4} ( 2 )} d u

= 7 \cdot \int \frac{ln ^{3} ( u )}{8 ln ^{4} ( 2 )} d u

Take the constant out:

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

= 7 \cdot \frac{1}{8 ln ^{4} ( 2 )} \cdot \int ln^{3} (u) d u

Apply Integration By Parts

: u ln^{3} (u) - \int 3 ln^{2} (u) d u

= 7 \cdot \frac{1}{8 ln ^{4} ( 2 )} (u ln^{3} (u) - \int 3 ln^{2} (u) d u)

\int 3 ln^{2} (u) d u = 3 (u ln^{2} (u) - 2 (u ln (u) - u))

= 7 \cdot \frac{1}{8 ln ^{4} ( 2 )} (u ln^{3} (u) - 3 (u ln^{2} (u) - 2 (u ln (u) - u)))

Substitute back

u = 4^{x}

= 7 \cdot \frac{1}{8 ln ^{4} ( 2 )} (4^{x} ln^{3} (4^{x}) - 3 (4^{x} ln^{2} (4^{x}) - 2 (4^{x} ln (4^{x}) - 4^{x})))

Simplify

7 \cdot \frac{1}{8 ln ^{4} ( 2 )} (4^{x} ln^{3} (4^{x}) - 3 (4^{x} ln^{2} (4^{x}) - 2 (4^{x} ln (4^{x}) - 4^{x}))) : \frac{7}{8 ln ^{4} ( 2 )} (ln^{3} (2) \cdot 2^{2 x + 3} x^{\frac{3}{2}} - 3 (ln^{2} (2) \cdot 4^{x + 1} x - 2 (ln (2) \cdot 2^{2 x + 1} x - 4^{x})))

= \frac{7}{8 ln ^{4} ( 2 )} (ln^{3} (2) \cdot 2^{2 x + 3} x^{\frac{3}{2}} - 3 (ln^{2} (2) \cdot 4^{x + 1} x - 2 (ln (2) \cdot 2^{2 x + 1} x - 4^{x})))

Add a constant to the solution

= \frac{7}{8 ln ^{4} ( 2 )} (ln^{3} (2) \cdot 2^{2 x + 3} x^{\frac{3}{2}} - 3 (ln^{2} (2) \cdot 4^{x + 1} x - 2 (ln (2) \cdot 2^{2 x + 1} x - 4^{x}))) + C

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