What is the integral from 0 to 1 of (x^2)/((x+3)^2) ?

The integral from 0 to 1 of (x^2)/((x+3)^2) is -6(2ln(2)-ln(3))+7/4

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integral from 0 to 1 of (x^2)/((x+3)^2)

\int_{0}^{1} \frac{x ^{2}}{( x + 3 ) ^{2}} d x

- 6 (2 ln (2) - ln (3)) + \frac{7}{4}

Decimal

0.02390 \dots

Solution steps

\int_{0}^{1} \frac{x ^{2}}{( x + 3 ) ^{2}} d x

Apply u-substitution

: \int_{3}^{4} \frac{( u - 3 ) ^{2}}{u ^{2}} d u

= \int_{3}^{4} \frac{( u - 3 ) ^{2}}{u ^{2}} d u

Expand

\frac{( u - 3 ) ^{2}}{u ^{2}} : 1 - \frac{6}{u} + \frac{9}{u ^{2}}

= \int_{3}^{4} 1 - \frac{6}{u} + \frac{9}{u ^{2}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{3}^{4} 1 d u - \int_{3}^{4} \frac{6}{u} d u + \int_{3}^{4} \frac{9}{u ^{2}} d u

\int_{3}^{4} 1 d u = 1

\int_{3}^{4} \frac{6}{u} d u = 6 (2 ln (2) - ln (3))

\int_{3}^{4} \frac{9}{u ^{2}} d u = \frac{3}{4}

= 1 - 6 (2 ln (2) - ln (3)) + \frac{3}{4}

Combine the fractions

1 + \frac{3}{4} : \frac{7}{4}

= - 6 (2 ln (2) - ln (3)) + \frac{7}{4}

\int e^{10 x} \cdot 5 x d x

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\int_{- 4}^{4} (2 x^{7} + 1) d x

(\frac{1}{y ^{2}} - \frac{9}{y ^{4}}) (y + 3 y^{3})

\int_{2}^{4} f (x) d x

What is the integral from 0 to 1 of (x^2)/((x+3)^2) ?
The integral from 0 to 1 of (x^2)/((x+3)^2) is -6(2ln(2)-ln(3))+7/4