What is the integral from 0 to 1 of sinh(4x)(x^2+5) ?

The integral from 0 to 1 of sinh(4x)(x^2+5) is (45e^8+53)/(64e^4)-41/32

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integral from 0 to 1 of sinh(4x)(x^2+5)

\int_{0}^{1} sinh (4 x) (x^{2} + 5) d x

\frac{45 e ^{8} + 53}{64 e ^{4}} - \frac{41}{32}

Solution steps

\int_{0}^{1} sinh (4 x) (x^{2} + 5) d x

Apply Integration By Parts

: [\frac{1}{4} cosh (4 x) (x^{2} + 5) - \int \frac{1}{2} x cosh (4 x) d x]_{0}^{1}

= [\frac{1}{4} cosh (4 x) (x^{2} + 5) - \int \frac{1}{2} x cosh (4 x) d x]_{0}^{1}

\int \frac{1}{2} x cosh (4 x) d x = \frac{1}{2} (\frac{1}{4} x sinh (4 x) - \frac{1}{16} cosh (4 x))

= [\frac{1}{4} cosh (4 x) (x^{2} + 5) - \frac{1}{2} (\frac{1}{4} x sinh (4 x) - \frac{1}{16} cosh (4 x))]_{0}^{1}

Simplify

[\frac{1}{4} cosh (4 x) (x^{2} + 5) - \frac{1}{2} (\frac{1}{4} x sinh (4 x) - \frac{1}{16} cosh (4 x))]_{0}^{1} : [\frac{1}{32} (8 x^{2} cosh (4 x) - 4 x sinh (4 x) + 41 cosh (4 x))]_{0}^{1}

= [\frac{1}{32} (8 x^{2} cosh (4 x) - 4 x sinh (4 x) + 41 cosh (4 x))]_{0}^{1}

Compute the boundaries:

\frac{45 e ^{8} + 53}{64 e ^{4}} - \frac{41}{32}

= \frac{45 e ^{8} + 53}{64 e ^{4}} - \frac{41}{32}

\int_{0}^{1} \frac{1}{x ^{2} - 2 x - 3} d x

\int_{- 1}^{4} x^{2} d x

\int_{0}^{3} (2 - x) d x

\int_{- 2}^{2} x^{3} + x d x

\int_{1}^{3} 3 x (x^{2} + 5) d x

What is the integral from 0 to 1 of sinh(4x)(x^2+5) ?
The integral from 0 to 1 of sinh(4x)(x^2+5) is (45e^8+53)/(64e^4)-41/32