Step-by-Step Calculus Solutions: Three Classic Definite & Indefinite Integration Problems

$u_1 = x^2 \implies du_1 = 2x \, dx, \quad dv_1 = e^x \, dx \implies v_1 = e^x$

$\int x^2 e^x \, dx = x^2 e^x - \int 2x e^x \, dx = x^2 e^x - 2 \int x e^x \, dx$

$u_2 = x \implies du_2 = dx, \quad dv_2 = e^x \, dx \implies v_2 = e^x$

$\int x e^x \, dx = x e^x - \int e^x \, dx = x e^x - e^x$

$\int x^2 e^x \, dx = x^2 e^x - 2(x e^x - e^x) + C = e^x \left( x^2 - 2x + 2 \right) + C$

$x = a \sin\theta \implies dx = a \cos\theta \, d\theta, \quad \theta = \arcsin\left(\frac{x}{a}\right)$

$\sqrt{a^2 - x^2} = \sqrt{a^2 - a^2 \sin^2\theta} = \sqrt{a^2(1 - \sin^2\theta)} = a \cos\theta$

$\int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \int \frac{a \cos\theta}{a \cos\theta} \, d\theta = \int 1 \, d\theta$

$\int 1 \, d\theta = \theta + C = \arcsin\left(\frac{x}{a}\right) + C$

$\frac{2x+3}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$

$2x + 3 = A(x+2) + B(x+1)$

$x = -1 \implies 1 = A(1) \implies A = 1; \quad x = -2 \implies -1 = B(-1) \implies B = 1$

$\int \left( \frac{1}{x+1} + \frac{1}{x+2} \right) dx = \ln|x+1| + \ln|x+2| + C$

$\int \frac{2x+3}{x^2+3x+2} \, dx = \ln\left| (x+1)(x+2) \right| + C = \ln\left| x^2+3x+2 \right| + C$