What is the integral from 1 to 5 of ln(t^2+5t+1) ?

The integral from 1 to 5 of ln(t^2+5t+1) is

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integral from 1 to 5 of ln(t^2+5t+1)

\int_{1}^{5} ln (t^{2} + 5 t + 1) d t

\frac{15 ln ( 51 ) + 21 ln ( \frac{5 3}{7} + 1 ) + 21 ln ( \frac{7}{3} - 1 ) - 16 - 7 ln ( 7 ) - 21 ln ( \frac{5 3}{7} - 1 ) - 21 ln ( \frac{7}{3} + 1 )}{2}

Decimal Notation

12.53418 \dots

Solution steps

\int_{1}^{5} ln (t^{2} + 5 t + 1) d t

Complete the square

t^{2} + 5 t + 1 : (t + \frac{5}{2})^{2} - \frac{21}{4}

= \int_{1}^{5} ln ((t + \frac{5}{2})^{2} - \frac{21}{4}) d t

Apply u-substitution

: \int_{\frac{7}{2}}^{\frac{15}{2}} ln (u^{2} - \frac{21}{4}) d u

= \int_{\frac{7}{2}}^{\frac{15}{2}} ln (u^{2} - \frac{21}{4}) d u

Apply Integration By Parts

: [u ln (u^{2} - \frac{21}{4}) - \int \frac{8 u ^{2}}{4 u ^{2} - 21} d u]_{\frac{7}{2}}^{\frac{15}{2}}

= [u ln (u^{2} - \frac{21}{4}) - \int \frac{8 u ^{2}}{4 u ^{2} - 21} d u]_{\frac{7}{2}}^{\frac{15}{2}}

\int \frac{8 u ^{2}}{4 u ^{2} - 21} d u = - \frac{21}{2} ln \frac{2}{21} u + 1 + \frac{21}{2} ln \frac{2}{21} u - 1 + 2 u

= [u ln (u^{2} - \frac{21}{4}) - (- \frac{21}{2} ln \frac{2}{21} u + 1 + \frac{21}{2} ln \frac{2}{21} u - 1 + 2 u)]_{\frac{7}{2}}^{\frac{15}{2}}

Simplify

[u ln (u^{2} - \frac{21}{4}) - (- \frac{21}{2} ln \frac{2}{21} u + 1 + \frac{21}{2} ln \frac{2}{21} u - 1 + 2 u)]_{\frac{7}{2}}^{\frac{15}{2}} : [\frac{1}{2} (2 u ln (u^{2} - \frac{21}{4}) + 21 ln \frac{2}{21} u + 1 - 21 ln \frac{2}{21} u - 1 - 4 u)]_{\frac{7}{2}}^{\frac{15}{2}}

= [\frac{1}{2} (2 u ln (u^{2} - \frac{21}{4}) + 21 ln \frac{2}{21} u + 1 - 21 ln \frac{2}{21} u - 1 - 4 u)]_{\frac{7}{2}}^{\frac{15}{2}}

Compute the boundaries:

\frac{15 ln ( 51 ) + 21 ln ( \frac{5 3}{7} + 1 ) + 21 ln ( \frac{7}{3} - 1 ) - 16 - 7 ln ( 7 ) - 21 ln ( \frac{5 3}{7} - 1 ) - 21 ln ( \frac{7}{3} + 1 )}{2}

= \frac{15 ln ( 51 ) + 21 ln ( \frac{5 3}{7} + 1 ) + 21 ln ( \frac{7}{3} - 1 ) - 16 - 7 ln ( 7 ) - 21 ln ( \frac{5 3}{7} - 1 ) - 21 ln ( \frac{7}{3} + 1 )}{2}

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

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

\int_{0}^{1} x^{2} arccos (x) d x

\int_{0}^{1} x ln (x^{2} + 1) d x

\int_{\frac{1}{2}}^{1} (2 x - 1)^{2019} d x

What is the integral from 1 to 5 of ln(t^2+5t+1) ?
The integral from 1 to 5 of ln(t^2+5t+1) is