How do you integrate $\int_{1}^{7} \sqrt{x} \cdot ln x$ $int_1^7 sqrtx*lnx$ using integration by parts?

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How do you integrate $\int_{1}^{7} \sqrt{x} \cdot ln x$ $int_1^7 sqrtx*lnx$ using integration by parts?

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Answer 1 · 2016-12-04T00:38:28

$\int_{1}^{7} \sqrt{x} ln x d x = \frac{2}{3} \sqrt{343} (ln 7 - \frac{2}{3}) + \frac{4}{9}$ $int_1^7sqrt(x)lnxdx = 2/3sqrt(343)(ln7 - 2/3)+4/9$

Explanation:

$\int_{1}^{7} \sqrt{x} ln x d x = \frac{2}{3} \int_{1}^{7} ln x d (x^{\frac{3}{2}})$ $int_1^7sqrt(x)lnxdx = 2/3int_1^7lnx d(x^(3/2))$

$\int_{1}^{7} \sqrt{x} ln x d x = \frac{2}{3} (x^{\frac{3}{2}} ln x) ∣_{x = 1}^{x = 7} - \frac{2}{3} \int_{1}^{7} x^{\frac{3}{2}} d (ln x)$ $int_1^7sqrt(x)lnxdx = 2/3(x^(3/2)lnx )|_(x=1)^(x=7) - 2/3int_1^7x^(3/2)d(lnx)$

$\int_{1}^{7} \sqrt{x} ln x d x = \frac{2}{3} (x^{\frac{3}{2}} ln x) ∣_{x = 1}^{x = 7} - \frac{2}{3} \int_{1}^{7} x^{\frac{3}{2}} \frac{d x}{x}$ $int_1^7sqrt(x)lnxdx = 2/3(x^(3/2)lnx )|_(x=1)^(x=7) - 2/3int_1^7x^(3/2)dx/x$

$\int_{1}^{7} \sqrt{x} ln x d x = \frac{2}{3} (x^{\frac{3}{2}} ln x) ∣_{x = 1}^{x = 7} - \frac{2}{3} \int_{1}^{7} x^{\frac{1}{2}} d x$ $int_1^7sqrt(x)lnxdx = 2/3(x^(3/2)lnx )|_(x=1)^(x=7) - 2/3int_1^7x^(1/2)dx$

$\int_{1}^{7} \sqrt{x} ln x d x = \frac{2}{3} x^{\frac{3}{2}} (ln x - \frac{2}{3}) {∣ ∣ ∣}_{x = 1}^{x = 7}$ $int_1^7sqrt(x)lnxdx = 2/3x^(3/2)(lnx - 2/3)|_(x=1)^(x=7)$

$\int_{1}^{7} \sqrt{x} ln x d x = \frac{2}{3} \sqrt{343} (ln 7 - \frac{2}{3}) + \frac{4}{9}$ $int_1^7sqrt(x)lnxdx = 2/3sqrt(343)(ln7 - 2/3)+4/9$

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How do you integrate $\int_{1}^{7} \sqrt{x} \cdot ln x$ $int_1^7 sqrtx*lnx$ using integration by parts?

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