Final Answer:
a) $\frac{1}{3}\ln|3x+7| + C$
b) $\frac{1}{3}x^3 e^{x^2} - \frac{1}{6}e^{x^2} + C$
c) $\frac{3}{8}x + \frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x + C$
d) $\frac{1}{4}x^4 - 2x^3 + 4x^2 + C$
e) $x^2 \arctan x - \frac{1}{2}x + \frac{1}{2}\arctan x + C$
f) $\frac{1}{6}\ln(3x^2 + 1) + \frac{1}{\sqrt{3}}\arctan(\sqrt{3}x) + C$
g) $\sin 3x + C$
h) $\frac{2}{5}x^{5/2} + C$
i) $-\frac{1}{2}e^{-x^2} + C$
j) $e^x - \frac{1}{3}e^{3x} + C$
k) $-\frac{15^x}{x\ln 15} + \frac{9^x}{x\ln 9} + C$ — *Wait, this is incorrect; correct approach below*
Actually, k) requires rewriting: $\int \frac{15^x - 9^x}{3^x} dx = \int \left(5^x - 3^x\right) dx = \frac{5^x}{\ln 5} - \frac{3^x}{\ln 3} + C$
l) $\frac{5^x}{\ln 5} + C$
m) $\frac{1}{6}\ln\left|\frac{x}{3x - 5}\right| + C$
n) $3\sin\left(\frac{x+2}{3}\right) + C$
o) $\int \frac{(3^x + 4^x)^2}{12^x} dx = \int \left(\left(\frac{3}{12}\right)^x + 2\left(\frac{12}{12}\right)^x + \left(\frac{4}{12}\right)^x\right) dx = \int \left((\tfrac{1}{4})^x + 2 + (\tfrac{1}{3})^x\right) dx = \frac{(1/4)^x}{\ln(1/4)} + 2x + \frac{(1/3)^x}{\ln(1/3)} + C$
p) $\ln|\ln x| + C$
q) $\frac{10^x}{x\ln 10} + \frac{4^x}{x\ln 4} + C$ — *Incorrect again*
Correct: $\int \frac{10^x + 4^x}{2^x} dx = \int \left(5^x + 2^x\right) dx = \frac{5^x}{\ln 5} + \frac{2^x}{\ln 2} + C$
r) $\int \frac{\sin 2x}{\cos x} dx = \int \frac{2\sin x \cos x}{\cos x} dx = \int 2\sin x\,dx = -2\cos x + C$
s) $\frac{(5x - 11)^{10}}{50} + C$
t) $\int \sec^2 x \csc^2 x\,dx = \int \frac{1}{\cos^2 x \sin^2 x} dx = \int \frac{4}{\sin^2 2x} dx = -2\cot 2x + C$
u) $-\cot x - x + C$
v) Let $u = \arcsin x$, then integral becomes $\int \frac{5 - 2u}{\sqrt{1 - u^2}} \cdot \frac{1}{\sqrt{1 - x^2}} dx$ — better: note derivative of $\arcsin x$ is $1/\sqrt{1-x^2}$, so let $u = \arcsin x$, $du = dx/\sqrt{1-x^2}$, then integral = $\int \frac{5 - 2u}{\sqrt{1 - u^2}} du$ — actually simpler: rewrite as $\int \left(5(1 - x^2)^{-1/2} - 2\arcsin x (1 - x^2)^{-1/2}\right) dx = 5\arcsin x - (\arcsin x)^2 + C$
w) $\frac{(1+x)^{16}}{16} + C$
x) $\int (-3x + 8)^{6/5} dx = -\frac{5}{33}(-3x + 8)^{11/5} + C$
y) Let $u = 10 + \sin x$, $du = \cos x\,dx$, so $\int \frac{du}{u} = \ln|10 + \sin x| + C$
z) $-\frac{1}{5}\cos(1 - 5x) + C$
Z) $-\frac{1}{3}e^{-3x + 1} + C$
Parent Tip: Review the logic above to help your child master the concept of indefinite integral worksheet.
